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RMM-ABSTRACT-ALGEBRA-MARATHON-101-200

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R M M
ROMANIAN MATHEMATICAL MAGAZINE
Founding EditorFounding Editor
DANIEL SITARUDANIEL SITARU
Available onlineAvailable online
www.ssmrmh.rowww.ssmrmh.ro
ISSN-L 2501-0099ISSN-L 2501-0099
RMM - Abstract Algebra Marathon 101 - 200RMM - Abstract Algebra Marathon 101 - 200
 
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 � RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Proposed by 
Daniel Sitaru – Romania,Marian Ursărescu-Romania, Ovidiu Pop-Romania, 
Angel Plaza-Spain, Moubinool Omarjee-France,Srinivasa Raghava-AIRMC-
India, Florentin Vişescu-Romania,Florică Anastase-Romania 
Rajeev Rastogi-India, Marin Chirciu-Romania,Rahim Shahbazov-Baku-
Azerbaijan, Dorin Mărghidanu-Romania,Nguyen Van Canh-Ben Tre-Vietnam 
Jalil Hajimir-Toronto-Canada,Costel Florea-Romania 
Rovsen Pirguliyev-Sumgait-Azerbaijan,H.Tarverdi-Baku-Azerbaijan 
Mihaly Bencze-Romania,Ionuţ Florin Voinea-Romania 
D.M. Bătineţu-Giurgiu – Romania,Neculai Stanciu-Romania 
Adil Abdullayev-Baku-Azerbaijan, Pavlos Trifon-Greece, Jhoaw Carlos-Brazil 
Mokhtar Khassani-Mostaganem-Algerie 
 
 
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 � RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Solutions by 
Daniel Sitaru-Romania,Florentin Vişescu-Romania,Ravi Prakash-New Delhi-
India,Khanh Hung Vu-Ho Chi Minh-Vietnam,Abdul Hannan-Tezpur-India 
Marian Ursărescu-Romania,Naren Bhandari-Bajura-Nepal,Sanong Huayrerai-
Nakon Pathom-Thailand,Florică Anastase-Romania,Oyebamiji Oluwaseyi-
Nigeria,George Florin Şerban-Romania,Rahim Shahbazov-Baku-Azerbaijan 
Santos Martins Junior-Brusels-Belgium,Abner Chinga Bazo-Peru 
Rovsen Pirguliyev-Sumgait-Azerbaijan,Adrian Popa-Romania 
Tran Hong-Dong Thap-Vietnam,Precious Itsuokor-Nigeria,Nas Meister-
Thailand,Ertan Yildirim-Turkey,Eldeniz Hesenov-Azerbaijan,Dorin 
Mărghidanu-Romania,Michael Sterghiou-Greece,Israel Castillo Pico-Brazil 
Sudhir Jha-Kolkata-India,Abdul Aziz-Semarang-Indonesia,Nikos Ntorvas-
Greece, Vivek Kumar-India,Khaled Abd Imouti-Damascus-Syria,Pavlos Trifon-
Greece, Bedri Jajrizi-Mitrovica-Kosovo,Izumi Ainsworth-Lima-Peru,Djeeraj 
Badera-Jaipur-India,Orlando Irahola Ortega-La Paz-Bolivia 
Lety Sauceda-Mexico City-Mexico,Arslan Ahmed-Sanaa-Yemen,Fayssal 
Abdelli-Bejaia-Algerie,Asmat Qatea-Afghanistan,Carlos Paiva-Brazil 
Christos Tsifakis-Greece,Do Chinh-Vietnam,Carlos Eduardo Aguiar Paiva-Brazil 
Agayev Seddredin-Azerbaijan,Rajeev Rastogi-India,Remus Florin Stanca-
Romania,Jalil Hajimir-Toronto-Canada,Jhoaw Carlos-Brazil 
Hussain Reda Zadah-Afghanistan 
 
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 � RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
101. �, � ∈ �����(ℝ), � ∈ ℝ − {�}, 
����� + ��(� + �) + ��
�(�� + ��) = �����. 
Find: Ω = ���(�� − ��) 
Proposed by Marian Ursărescu-Romania 
Solution by Florentin Vişescu-Romania 
����� + ��(� + �) + ��
�(�� + ��) = ����� ⇔ 
������ + ��(� + �) + ��
�(�� + ��) = ����� ⇔ 
(����� + ���)
� + (����� + ���)
� = ����� 
Let: � = ����� + ��� and � = ����� + ���, �, � ∈ �����(ℝ) then �
� + �� = ����� 
(� + ��)(� − ��) = �� − ��� + ��� + �� = �(�� − ��) ⇒ 
���(� + ��)(� − ��) = ���(� + ��)���(� − ��) = ���(� + ��)���(� + ��)���������������� ≥ � 
⇒ ���[�(�� − ��)] ≥ � ⇒ ��������(�� − ��) ≥ � 
⇒ �����(�� − ��) ≥ � ⇒ ���(�� − ��) = � 
�� − �� = (����� + ���)(����� + ���) − (����� + ���)(����� + ���) 
= ����� + ��� + ��� + ��
��� − ����� − ��� − ��� − ��
��� 
= ���(�� − ��) 
���(�� − ��) = ���[���(�� − ��)] = (−���)�������(�� − ��) ⇒ 
Ω = ���(�� − ��) = � 
 
102. Prove that: 
�
�������� ����
���� �
����� �
�������� ����
����� ��������
� ����
� �����
����� ��������
� ≠ �; ∀�, � ∈ ℝ 
Proposed by Daniel Sitaru-Romania 
 
 
 
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 � RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Solution by Ravi Prakash-New Delhi-India 
Let: ∆= �
�������� ����
���� �
����� �
�������� ����
����� ��������
� ����
� �����
����� ��������
� 
����� �� → �� − (����)�� ��� �� → �� − (����)�� 
∆= ��
 
� ����
� �
 � �
 � ����
����(� − ��������) ��������
� − �������� ����
�(� − �����) �����
����(� − ����� ��������
�� = 
= (� − ��������)(� − �����)∆� 
∆�= �
� ����
� �
� �
� ����
���� ��������
� ����
� �����
���� ��������
� 
����� �� → �� − (����)�� ��� �� → �� − (����)�� 
∆�= �
� ����
� �
 � �
 � ����
���� � 
� � 
� �
���� �
� 
����� �� → �� − (����)��, we get: 
∆�= �
� ����
� �
 � �
 � ����
� � 
� � 
� − �������� �
���� �
� =
���������
��
− �
���� � �
� � ����
� � − �������� �
� = 
=
���������
��
(� − ��������) �
���� �
� ����
� = 
= −(� − ��������)(� − �����)(� − ��������)� ≠ � 
103. � = �� �⁄ > �, �√� = � ∙ �√��; � = �� �⁄ > �, ��� = �� ∙ � ��
�
� 
� = �� �⁄ > �, �√� = ��� ∙ �� √�
�
�. Find the set Ω such that: 
�∆Ω∆� = �, (�∆� = (� �⁄ ) ∪ (� �⁄ ) 
Proposed by Daniel Sitaru-Romania 
 
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 � RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Solution by Khanh Hung Vu-Ho Chi Minh-Vietnam 
We have �√� = � ∙ �√� ⇒ �����√�� = ����� ∙ �√�� ⇒ 
√���� � = ���� + √����� ⇒ √���� � − ���� − √����� = �; (�) 
Put: ��(�) = √���� � − ���� − √�����; �
�
�
(�) =
�
�√�
���� �
�
�
� + �� 
��
�
(�) = � ⇔ ��� �
�
�
� + � = � ⇔ � =
�
��
 
We have: 
���
�→��
���(�)� = ���
�→��
�√���� � − ���� − √������ = ���
�→��
�
��� �
�
√�
� − ���� =
���
 
= ���
�→��
�
�
�
−�
�√�
� − ���� = ���
�→��
�−�√�� − ���� = −���� 
���
�→�
���(�)� = ���
�→�
�√���� � − ���� − √������ = ���
�→��
�√���� �
�
�
� − ����� = +∞ 
So, the equation (1) has only one root, which is � = �, then � = {�}. 
Since � ∉ � and since �√� = �� and ��� ∙ �� √�
�
∉ ℕ, so not have exist the set Ω such that 
�∆Ω∆� = � 
104. Let (�, +,∙) be a field with property – � = ���, ∀� ∈ �, � ≠ � prove that: 
(�, +,∙) ≃ (ℤ�, +,∙) 
Proposed by Ovidiu Pop-Romania 
Solution 1 by Ravi Prakash-New Delhi-India 
� = ��� = −� ⇒ � = � = � 
Characteristic of field � is 2, then no of elements in � is ��, � ∈ ℕ. 
Suppose, � ≥ �, let �� = � − {�}, then the number of elements in � is �
� − �. 
Also, �� −is a group under multiplication. 
Let � ∈ ��, � ≠ � then �
�� = −� = �, (�� = �) 
Not possible. Thus, � = �. 
 
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 � RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
��-is a field containing two elements. 
But here is one and only one field containing exactly two elements. 
Thus, (�, +,∙) ≃ (ℤ�, +,∙) 
Solution 2 by Abdul Hannan-Tezpur-India 
– � = ���, ∀� ∈ �, � ≠ � 
Put � = �: � = ��� = −� ⇒ � = � = �; (∗) 
Now, let � ∈ � − {�} be an arbitrary element. 
Then – ��� = � and hence, �� = −� ∙ ��� = −� =
(∗)
� 
�� − � = � ⇒ (� − �)(� + �) = � ⇒ (� − �)(� − �) = � using (*) again. 
Therefore, (� − �)� = � ⇒ � − � = � since � is a field ⇒ � = � ⇒ � = {�, �} 
Hence, (�, +,∙) ≃ (ℤ�, +,∙) with the isomorphism given by 
∅: � → ℤ�, ∅(�) = �� and ∅(�) = �.� 
 
105. Let �(�) = ���
� + ���
��� + ⋯ + ��, where ��, ��, … , �� are integers. 
Show that if � takes the value ���� for four distinct integral values of �, 
then � cannot take the value ���� for any integral value of �. 
Proposed by Angel Plaza-Spain 
Solution by Marian Ursărescu-Romania 
By absurdum: let’s say ∃� ∈ ℤ such that �(�) = ����; (�) 
Let �(�) = �(�) − ����, from hypothesis ∃��, ��, ��, �� ∈ ℤ (distinct) such that 
�(��) = �(��) = �(��) = �(��) = ���� ⇒ 
⎩
⎨
⎧
�(��) = �
�(��) = �
�(��) = �
�(��) = �
⇒ �
� − �� ∣ �
� − �� ∣ �
� − �� ∣ �
� − �� ∣ �
⇒ (� − ��)(� − ��)(� − ��)(� − ��) ∣ � 
�(�) = (� − ��)(� − ��)(� − ��)(� − ��)�(�) ⇒ 
�(�) − ���� = (� − ��)(� − ��)(� − ��)(� − ��)�(�); (�) 
From (1),(2) we get: 
 
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 � RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
−�� = (� − ��)(� − ��)(� − ��)(� − ��)�(�), relation which it is obviously false. 
 
106. � real polynomial degree � ≥ � such that 
�(�), �(�), �(�), �(�), … , �(��) are in ℤ. Prove that ∀� ∈ ℤ, �(��) ∈ ℤ. 
Proposed by Moubinool Omarjee-France 
Solution by Abdul Hannan-Tezpur-India 
Lemma. Let�(�) be a real polynomial of degree � such that 
�(�), �(�), �(�), … , �(�) ∈ ℤ . Then �(�) ∈ ℤ, ∀� ∈ ℤ. 
Before proving the lemma, let us see why it solves the given question. 
Let �(�) be a real polynomial of degree � such that 
�(�), �(�), �(�), �(�), … , �(��) ∈ ℤ 
Define �(�) ≔ �((� − �)�). Then �(�) is a real polynomial of degree �� such that 
�(�), �(�), �(�), … , �(��) ∈ ℤ 
Hence by the lemma above, �(�) ∈ ℤ, ∀� ∈ ℤ ⇒ �(��) = �(� + �) ∈ ℤ, ∀� ∈ ℤ. 
Proof the lemma: We will prove this by induction on �. 
Base case: � = �: Let �(�) = �� + �. Then �(�) = � ∈ ℤ and �(�) = � + � ∈ ℤ 
This implies that both � and � are integers. So, ∀� ∈ ℤ, �(�) = �� + � ∈ ℤ 
This proves the base case. Assume now that the result is true for all polynomials of 
degree ≤ � (where � ≥ �) satisfying the given conditions. 
Induction step: Let degree of �(�) be � + � such that 
�(�), �(�), �(�), … , �(� + �) ∈ ℤ 
Define �(�) ≔ �(� + �) − �(�). Then �(�) is a polynomial of degree � such that 
�(�), �(�), �(�), … , �(�) ∈ ℤ 
Thus by induction hypothesis, �(�) ∈ ℤ, ∀� ∈ ℤ; (∗) 
If � > � + �, then �(�) = �(� + �) + ∑ �(�)�������� ∈ ℤ using (∗) 
If � < �, then �(�) = �(�) − ∑ �(�) ∈ ℤ����� using (∗) again 
⇒ �(�) ∈ ℤ, ∀� ∈ ℤ. This completes the proof of the lemma. 
 
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 � RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
107. If � ∈ ��(ℝ), ���� = �, ���(�
� + � + ��) = � then find: 
٠= ��� 
Proposed by Marian Ursărescu-Romania 
Solution 1 by Ravi Prakash-New Delhi-India 
���(�� + � + ��) = � ⇒ ���(� − ���)(� + �
���) = � 
���(� − ���)(� − ��������������) = ���(� − ���) ∙ ���(� − ���)������������������ 
Hence, |���(� − ���)|
� = � ⇒ ���(� − ���) = � 
�� − (���)�� + (���∗)� − � = � 
(���) �−
�
�
−
√�
�
�� = (���∗) �−
�
�
+
√�
�
�� 
��� = ���∗ and – ��� = ���∗. Therefore, ��� = ���∗ = � 
Solution 2 by Florentin Vişescu-Romania 
���(�� + � + ��) = ���(� − ���)(� − �� ��) = � 
��(�) = (� − �)(�
� + � + �) = �� − (���)�� + (���∗)� − ���� 
�� + (� − �)�� + (� − �)� − � = �� − (���)�� + (���∗)� − ���� 
� − � = −��� and � − � = ���∗ hence � = �. 
Therefore, ��� = ���∗ = � 
 Solution 3 by Adrian Popa-Romania 
���(�� + � + ��) = ���(� − ���)(� − ����) = �, �
� = � 
��(�) = (� − �)(� − �)(� + �) = �
� − (���)�� + (���∗)� − ���� 
�� + (� − �)�� + (� − �)� − � = �� − (���)�� + (���∗)� − ���� 
� − � = −��� and � − � = ���∗ hence � = �. 
Therefore, ��� = ���∗ = � 
 
108. If � ∈ ��(ℝ), ����
∗ = −�, ���(�� + � + ��) = � then find: 
٠= ���� 
Proposed by Marian Ursărescu-Romania 
 
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 � RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Solution 1 by Ravi Prakash-New Delhi-India 
���(�� + � + ��) = � ⇒ ���(� − ���)(� + �
���) = � 
���(� − ���)(� − ��������������) = ���(� − ���) ∙ ���(� − ���)������������������ 
Hence, |���(� − ���)|
� = � ⇒ ���(� − ���) = � 
�� − (���)�� + (���∗)� − ���� = � 
� − (���)�� − � − ���� = � 
−
√�
�
��� +
√�
�
= � 
Hence ��� = �. Therefore, ���� = � 
 Solution 2 by Florentin Vişescu-Romania 
���(�� + � + ��) = ���(� − ���)(� − �� ��) = � 
��(�) = (� − �)(�
� + � + �) = �� − (���)�� + (���∗)� − ���� 
�� + (� − �)�� + (� − �)� − � = �� − (���)�� + (���∗)� − ���� 
� − � = −��� and � − � = −� hence � = �, ��� = �. 
Therefore, ���� = � 
109. If � =
�
��
�
√��
�
��
�
√�
�⋯
 then evaluate this continued fraction 
� +
�
� +
�
� + � +
�
� + � +
�
� + � + ⋯
 
Proposed by Srinivasa Raghava-AIRMC-India 
Solution by Naren Bhandari-Bajura-Nepal 
Note that initial continued fraction can be written as: 
� = 
�
��
�
√��
�
��
�
√�
�⋯
=
�
��
�
√���
⇒ � =
����√��
√�����
 
From the last equation we have: �� + �√� − �√� = �. 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Solving the quadratic equation we yield 
� =
−√� + �� + ��√�
�
=
�
�
���� + �√� − √�� 
Let � =
�
��
�
����⋯
= � +
�
��
�
�
= � +
��
����
=
�������
����
 
Simplifying and solving quadratically gives 
��� − �� − � = � ⇒ � =
� + ��� + ���
��
=
�
�
�� + �−� + √� + ��� − ��√�� 
110. If � ∈ ��(ℝ), � = �
�, ��(��) = � �
���
�
�
�
. Prove that: 
��(��) = � �
���
�
�
�
, � ∈ ℕ, � ≥ � 
Proposed by Florentin Vişescu-Romania 
Solution by Marian Ursărescu-Romania 
� ∈ ��(ℝ), � = �
� ⇒ the eigenvalues of � are real⇒ ��, ��, �� ∈ ℝ 
���� = � �
���
�
�
�
⇔ ��
� + ��
� + ��
� =
�
�
(�� + �� + ��)
� ⇔ 
��
� + ��
� + ��
� = ���� + ���� + ���� ⇔ �� = �� = �� = � ⇒ 
The conclusion is obvious: ���� = ��
� + ��
� + ��
� = ��� and 
� �
���
�
�
�
= � �
��
�
�
�
= ���, � ∈ ℕ∗, its true for � = �, � = �. 
 
111. If � ≥ �, �, � > � then: 
����
��
+
�����
�� + ��
≥ ��
���� + �����
���� + �����
 
Proposed by Daniel Sitaru-Romania 
Solution by Sanong Huayrerai-Nakon Pathom-Thailand 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Iff �
����
��
+
�����
�����
� �
����
��
+
�����
�����
� (���� + �����) ≥ �(���� + �����) 
(���� + �����)�
������� + ����(�� + ��)�
∙
(���� + �����)�
������� + ����(�� + ��)�
∙ (���� + �����) ≥ �(���� + �����) 
(���� + �����)�(���� + �����) ≥ �(������ + ������ + �����)� 
��
���������������
� + ��
���������������
� �
�
≥ �(������ + ������ + �����)� 
����
�
� + ����
�
��
�
≥ �(������ + ������ + �����) 
����� + ������ + ����
�
����
�
� ≥ �(������ + ������) 
����
��
+
����
��
+ ��
�
��
�
� ≥ �(� + �) 
��(���)
���
+
��(���)
���
+ ��� ≥ �(�� + ��); � = ��, � = �� 
��(���)
��(���)
+
��(���)
��(���)
+ ��� ≥
��
��
+
��
��
+ ��� ≥ �(�� + ��) 
True because ∀� ≥ �, ∃� ∈ ℝ� such that � = � + �. 
 
112. If � < � ≤ � < � then: 
��� �
����
����
� ≥ �� + �
�
�
� ��� �
����
����√���
� 
Proposed by Daniel Sitaru-Romania 
Solution by Florică Anastase-Romania 
��� �
����
����
� ≥ �� + �
�
�
� ��� �
����
����√���
� 
⇔ ��� �
����
����
� − ��� �
����
����√���
� ≥ �
�
�
��� �
����
����√���
� 
⇔ ��� �
���√��
����
� ≥ �
�
�
��� �
����
����√���
� 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
⇔ √��������√��� − √����(����) ≥ √����(����) − √��������√��� 
⇔ �√� + √���������√��� ≥ √����(����) + √����(����) 
⇔ �������√��� ≥
√�
√� + √�
���(����) +
√�
√� + √�
���(����); (�) 
We have: 
√�
√��√�
���(����) +
√�
√��√�
���(����) ≤
������������
 
≤ ��� �
√�
√� + √�
���� +
√�
√� + √�
����� ≤
������������ (�,�)
 
≤ ��� ���� �
�√� + �√�
√� + √�
�� = ��� �����√���� ; (�) 
From (1),(2) it follows that: ��� �
����
����
� ≥ �� + �
�
�
� ��� �
����
����√���
� 
 
113. If � < � < � < � and �, �, � > � 
� = �� + ��
�
�
� �� + ��
�
� − �
� + �� + ��
�
�
 � �� + ��
�
� − �
� + �� + ��
�
�
� �� + ��
�
� − �
� 
Ω� = �� + ��√����
�
�
�
 
Prove that: Ω� ≥ �Ω� 
Proposed by Florică Anastase-Romania 
Solution 1 by Oyebamiji Oluwaseyi-Nigeria 
Given � < � < � < � 
� + ��
�
�
> 1 + ��
�
�
= � + � 
� + ��
�
� − �
> 1 + ��
�
�
= � + � 
The same applies for all the analogs. 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Ω� > (� + �)(� + �) + (� + �)(� + �) + (� + �)(� + �) = 
[� + (� + �) + ��] + [� + (� + �) + ��] + [� + (� + �) + ��] ≥
���������
 
≥
��� + (� + �) + �� + �� + (� + �) + �� + �� + (� + �) + ���
�
� + � + �
≥ 
But �� + (� + �) + �� ≥
���������
� +
���
�
+
��
�
 and analogs. 
≥
�� +
�� + �� + �� + �� + �� + ��
� �
�
�
= 
=
� �� +
�(� + � + �) + �� + �� + ��
� �
�
�
≥
���
 
= � �� + ��√����
�
�
�
= �Ω� 
 Solution 2 by George Florin Şerban-Romania 
�� + ��
�
�
� �� + ��
�
� − �
� ≥
�������
�� + ���
�
�
∙ ��
�
� − �
�
�
= 
= �� + �
���
��(� − �)
�
�
≥
���
�� + �
���
� + � − �
�
�
�
= �� + √����
�
 
Similarly: 
�� + ��
�
�
 � �� + ��
�
� − �
� ≥ �� + √����
�
 
�� + ��
�
�
� �� + ��
�
� − �
� ≥ �� + √����
�
 
��� + √����
�
���
≥ � �� + ��√����
�
�
�
⇔ 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
�
�
��� + √����
�
���
≥ �� + ��√����
�
�
�
= �� + �√��� ∙ √��� ∙ √���
�
�
�
 
Denote: √��� = �; √��� = �; √��� = � it enough to prove that:�
�
�(� + �)�
���
≥ �� + ����
�
�
�
 
�
�
�(� + �)�
���
≥
���
��(� + �)
�
���
� ⇒ 
��(� + �)
�
���
� ≥ �� + ����
�
�
�
 
��(� + �)
���
� ≥ � + ����
�
⇔ �(� + �)
���
≥ �� + ����
�
�
�
(�������) 
Therefore, Ω� ≥ �Ω� 
Solution 3 by Abdul Hannan-Tezpur-India 
�� + ��
�
�
� �� + ��
�
� − �
� ≥
���
�� + √�√��
�
��(� − �)
�
�
≥
���
 
≥ �� + √���
��
� + (� − �)
�
�
= �� + √����
�
 
Equality holds if � = � and � = ��. 
Similarly, we obtain the analogous with equality if � = �, � = �� and � = �, � = ��. 
If � = ��. � = �� then � = � false. Therefore, 
� �� + ��
�
�
� �� + ��
�
� − �
� >
���
��� + √����
�
���
≥
���
 
≥
���
���� + √����
�
�� + √����
�
�� + √����
��
= 
= � ��� + √����
�
��� + √����
�
��� + √����
�
��
�
≥
�������
 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
≥
�������
� �� + �√���√���√����
�
��
�
= � �� + ��√����
�
�
�
 
 
114. If � ≤ � ≤ � then: 
� �������
��
���
≤ ����� + ����(� − �) 
Proposed by Daniel Sitaru-Romania 
Solution by Rahim Shahbazov-Baku-Azerbaijan 
� �������
��
���
≤ ����� + ����(� − �); (�) 
�
�
= � ≥ �
(�)
�� ��� + �� + ⋯ + � + � ≤ �� + ����(� − �) 
If � = � the problem is solved. Suppose � > � → � − � > �. By multiplying: 
��� − � ≤ ��(� − �) + ����(� − �)� 
����� − ������ + ���� + ��� − �� ≥ � 
(� − �)������ + ���� + ���� + ���� + ���� + ���� + ���� + ��� + ��� ≥ � 
 
115. If �, � ∈ ℝ are such that: 
��� + �� − �� + � + ��� + �� + � = ��� + �� − �� − � + ��� + �� + �� − � 
then find: ٠= �� + � 
Proposed by Rajeev Rastogi-India 
Solution 1 by Santos Martins Junior-Brusels-Belgium 
Let �� + �� − �� + � = ��; (�) 
�� + �� + � = ��; (�) 
�� + �� − �� − � = ��; (�)�� + �� + �� − � = ��; (�), where �, �, �, � ≥ � 
Hence equation becomes: � + � = � + �; (�) 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
⇔ � − � = � − � ⇔ �� + �� − ��� = �� + �� − ���; (�) 
Also observe that (�) + (�) = (�) + (�) ⇔ �� + �� = �� + �� 
Using (�) in (�) we get: −��� = −��� ⇔ �� = ��; (�) 
Doing (�) + � ∙ (�): (� + �)� = (� + �)� ⇔ �(� + �) − (� + �)�(� + � + � + �) = � 
�(� − �)�(� + �) + (� + �)� = � 
Either: � = � ⇒ �� = �� ⇔ �� + �� − �� + � = �� + �� − �� − � 
⇔ �� + � = � 
Or � + � + � + � = � but since �, �, �, � ≥ � we must have: � = � = � = � = � meaning 
we have to determine couples (�, �) such that � = � = � = � = � holds simultaneously, 
hence �� = �� = �� = �� = � ⇒ �� + �� + �� + �� = � 
⇔ ��� + �� = � ⇔ �(� + �) = � ⇔ � = � or � = −� but neither of the two values of 
�, � such that � = � = � = � = � holds simultaneously. 
Hence, ٠= �� + � = � 
Solution 2 by Abner Chinga Bazo-Peru 
��� + �� − �� + � + ��� + �� + � = ��� + �� − �� − � + ��� + �� + �� − � 
��� + �� − �� + � − ��� + �� − �� − � = ��� + �� + �� − � − ��� + �� + � 
��������������������������������������������������
�����������������������
= 
=
��������������������������������������������
��������������������
 
−(�� + � − �)
��� + �� − �� + � + ��� + �� − �� − �
=
�� + � − �
��� + �� + �� − � + ��� + �� + �
 
�� + � − �. Hence, Ω = �� + � = � 
 
116. If � ∈ ℂ, � ∈ [�, ��), |�| = � then: 
��� + � − ���� + ������ + ����� + ����� − �� + �� + ����� + ����� + �� − �� ≤ � 
Proposed by Daniel Sitaru-Romania 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Solution by Rovsen Pirguliyev-Sumgait-Azerbaijan 
Denote: �� = �
�, �� = �, �� = ���� + ����� ��� �� + �� + �� = �� then we have: 
��� = |�� − ���| + |�� − ���| + |�� − ���|; (�) 
Now, applying Cauchy-Schwartz inequality, we get: 
(|�� − ���| + |�� − ���| + |�� − ���|)
� ≤ ��|�� − ���|
� + |�� − ���|
� + |�� − ���|
��; (�) 
�|�� − ���|
�
�
���
= �|��|
� − �������� − �������� + �(|��|
� + |��|
� + |��|
�) = −|��|
� + �� 
Hence,(|�� − ���| + |�� − ���| + |�� − ���|)
� ≤ �(−|��|
� + ��) ≤ �� ⇔ 
��� + � − ���� + ������ + ����� + ����� − �� + �� + ����� + ����� + �� − �� ≤ � 
 
117. ��, ��, �� ∈ ℂ − {�}, different in pairs, 
|��| = |��| = |��| = �, �(��), �(��), �(��) 
� �
(�� − ��)(�� − ��)
��� − �� − ��
�
���
= � ⇒ �� = �� = �� 
Proposed by Marian Ursărescu-Romania 
Solution 1 by Rovsen Pirguliyev-Sumgait-Azerbaijan 
Denote �(��), �(��), �(��), then �� = |�� − ��|, �� = |�� − ��|, �� = |�� − ��| 
� �
(�� − ��)(�� − ��)
��� − �� − ��
�
���
= � �
(�� − ��)(�� − ��)
(�� − ��) + (�� − ��
�
���
= �
�� ∙ ��
�� + ��
���
= �; (�) 
Since, 
��
���
≤
���
�
, then 
� = �
�� ∙ ��
�� + ��
���
≤
�� + �� + ��
�
⇒ �� + �� + �� ≥ � ⇒ 
|�� − ��| + |�� − ��| + |�� − ��| ≥ �; (�) 
Using: |�� − ��| ≤ |��| + |��| we have�
|�� − ��| ≤ �
|�� − ��| ≤ �
|�� − ��| ≤ �
⇒ 
|�� − ��| + |�� − ��| + |�� − ��| ≤ �; (�) 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
From (2),(3) we get: |�� − ��| + |�� − ��| + |�� − ��| = � 
Equality holds if |�� − ��| = |�� − ��| = |�� − ��| ⇒ �� = �� = ��. 
Solution 2 by Adrian Popa-Romania 
 
|��| = |��| = |��| = � ⇒ �(��), �(��), �(��) ∈ �(�, �) 
� < �, � < �, � < � ⇒ � + � + � < � 
|�� − ��| = ���������⃗ − ��������⃗ � = ���������⃗ � = � 
Similarly: |�� − ��| = �, |�� − ��| = � 
|��� − �� − ��| = |�� − �� + �� − ��| = ��������⃗ − �������⃗ + �������⃗ − �������⃗ � = 
= ���������⃗ + �������⃗ � = ���′�������⃗ � = ��� 
��
���
=
��
���
=
��
���
= � 
�
��� < � + �
��� < � + �
��� < � + �
⇒ � ≥
��
� + �
+
��
� + �
+
��
� + �
; (�) 
But 
��
���
<
���
�
 and analogs⇒ ∑
��
���
< ∑
���
�
=
�(�����)
�
=
�����
�
≤
�
�
= �; (�) 
From (1),(2) we get: 
��
� + �
+
��
� + �
+
��
� + �
= � 
��
� + �
+
��
� + �
+
��
� + �
≤
� + �
�
+
� + �
�
+
� + �
�
=
� + � + �
�
 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
� <
�����
�
⇒ � + � + � > �, but � + � + � < � then � + � + � = �. 
So, |�� − ��| + |�� − ��| + |�� − ��| = � 
�
|�� − ��| ≤ |��| + |��|
|�� − ��| ≤ |��| + |��|
|�� − ��| ≤ |��| + |��|
 
Equality holds if |�� − ��| = |�� − ��| = |�� − ��| ⇒ �� = �� = ��. 
 
118. (�� + � − �)� + (��� + �� + �)� + (��� − � − �)� = �, � ∈ ℂ 
Find: Ω = |�| 
Proposed by Daniel Sitaru-Romania 
Solution by Santos Martins Junior-Brusels-Belgium 
Let: �
� = �� + � − � = (� − �)(� + �) = � ∙ �
� = ��� + �� + � = (�� + �)(� + �) = � ∙ �
� = ��� − � − � = (� − �)(�� + �) = � ∙ �
 
Equation becomes: ���� + ���� + ���� = � 
(�� + �� + ��)� = ����(� + � + �); (�) 
Also, � + � = �; (�) 
(1) Can be rewritten: 
(�(� + �) + ��)� = ���(� + �)��(� + �)� 
((� + �)� + ��)� = ���(� + �)� 
(� + �)� + ���� + ���(� + �)� = ���(� + �)� 
(� + �)� − ���(� + �)� + ���� = � 
((� + �)� − ��)� = � 
(� + �)� = � ⇔ �(� − �) + (� + �)�
�
= (� − �)(� + �) ⇔ 
(�� + �)� = (�� + � − �) ⇔ �(�� + � + �) = � ⇔ 
⎩
⎪
⎨
⎪
⎧
�� = −
�
�
+
�√�
�
�� = −
�
�
−
�√�
�
⇒ |��| = |��| = �. 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
119. If (�� + ���)
���� = �, � ∈ �, ���������, ��, �� ∈ ℝ then find � ∈ ℕ such that: 
�
� − �� + ���
� + �� + ���
��
���
= ���� 
Proposed by Daniel Sitaru-Romania 
Solution by Ravi Prakash-New Delhi-India 
Let �� = �� + ���, then |��|
���� = �(�� + ���)
����� = � ⇒ |��| = � 
�� ∙ ��� = � ⇒ ��� =
�
��
 
We assume �� ≠ � for all � ≤ � ≤ ��, so that �� ≠ �,for all � ≤ � ≤ ��. 
Now, 
��������
��������
=
�����
����
=
����
��(����)
=
�
����
−
�
��
=
�
����
− ��� 
As �, ��, ��, … , ��� −are roots of the equation �
���� = �, then 
� + �� + �� + ⋯ + ��� = � 
Hence, ��� + ��� + ⋯ + ���� = −� 
Also 
�
����
+
�
����
+ ⋯ +
�
�����
=
(����)���
�������
 
Put � = −� we get: 
−
�
�
− �
�
� + ��
+
�
� + ��
+ ⋯ +
�
� + ���
� =
(�� + �)(−�)��
(−�)���� − �
 
Therefore, 
�
����
+
�
����
+ ⋯ +
�
�����
=
����
�
−
�
�
= � 
�
� − �� + ���
� + �� + ���
��
���
= �� + � ⇒ �� + � = ���� ⇒ � = ���� 
120. �, �, �, � ∈ ℂ different in pairs, 
|�| = |�| = |�| = |�| = �, �(�), �(�), �(�), �(�). Prove that if: 
|� + �|� + |� + �|� + |� + �|� + |� + �|� + |� + �|� + |� + �|� = � 
then���� −rectangle 
Proposed by Florentin Vişescu-Romania 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Solution by Marian Ursărescu-Romania 
We have (� + �)��� + ��� + (� + �)(�� + ��) + (� + �)��� + ��� + (� + �)��� + ��� + 
+(� + �)��� + ��� + (� + �)��� + ��� = � 
(∴ |�| = � ⇔ |�|� = � ⇔ � ∙ �� = �) 
�� + ��� + ��� + ��� + ��� + ��� + ��� + ��� + ��� + ��� + ��� + ��� + ��� = � 
⇔ (� + � + � + �)��� + �� + �� + ��� = � 
⇔ (� + � + � = �)�� + � + � + �������������������� = � 
⇔ |� + � + � + �|� = � ⇔ � + � + � + � = �; (�) 
Lemma: If ��, ��, ��, �� ∈ ℂ such that |��| = |��| = |��| = |��| = � 
and �� + �� + �� + �� = � then ���� −is rectangle. 
From (1), lemma and |�| = |�| = |�| = |�| = � then ���� −is rectangle. 
Remark. Demonstration of the previous result (just in case) 
We have �� = |� − �|, �� = |� − �|, we want to show: 
�� = �� ⇔ |� − �| = |� − �| ⇔ |� − �|� = |� − �|� 
⇔ (� − �)��� − ��� = (� − �)��� − ��� ⇔ ��� − ��� − ��� = ��� − ��� − ��� 
⇔ ��� + ��� = ��� + ���; (�) 
 But � + � + � + � = � ⇔ � + � = −(� + �) ⇒ �� + �� = −(�� + ��) 
(� + �)��� + ��� = (� + �)��� + ��� ⇒ ��� + ��� + ��� = ��� + ��� + ��� 
��� + ��� = ��� + ���; (��) 
From (�), (��) ⇒ �� = �� 
Similarly, �� = �� hence ���� −inscriptible parallelogram. 
Therefore, ���� −is rectangle. 
121. If �, �, � > �, √�� + √�� + √�� = � then: 
� ��� + ��
���
+ � � �
��√��
� + �
���
≥ �√� 
Proposed by Daniel Sitaru-Romania 
 
www.ssmrmh.ro 
 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Solution by 1 Sanong Huayrerai-Nakon Pathom-Thailand 
For �, �, � > 0, √�� + √�� + √�� = � we give: � = ��, � = ��, � = �� 
� ��� + ��
���
+ � � �
��√��
� + �
���
= � ��� + ��
���
+ � � �
(��)�
�� + ��
���
≥ 
≥ ��� ���
�
+ �� ���
�
+ � � �
(��)�
�� + ��
���
≥ �√� 
Iff ∑ �� + �√� + ∑ �
(��)�
��������
≥ � ����. 
Because: 
�����
�
+ �√� ∙ �
(��)�
�����
≥ ���; (��� �������) 
 Solution 2 by Tran Hong-Dong Thap-Vietnam 
�� + �� ≥
�
�
(� + �)�; �, � > � ⇒ ��� + �� ≥
� + �
�
 
��� + �� + ��
��√��
� + �
= ��� + �� + ��
��√��
� + �
+ ��
��√��
� + �
≥ 
≥
� + �
√�
+ ��
��√��
� + �
+ ��
��√��
� + �
≥
���
��
�(��)�
√�
�
= �����√�
�
 
⇒ ��� + �� + ��
��√��
� + �
≥ �����√�
�
; (��� �������) 
Hence, 
� ��� + ��
���
+ � � �
��√��
� + �
���
≥ ���√�
�
� ��
���
≥
� ∑ ���(∑ �)�
���√�
�
∙
�∑ √���
�
�
= 
= ��√�
�
∙ �� = �√�. With: √�� = �; √�� = �; √�� = � 
Solution 3 by Precious Itsuokor-Nigeria 
� ��� + ��
���
+ � � �
��√��
� + �
���
≥
���
√� � √��
���
+
�
√�
� √��
���
= √� ∙ � + �√� ∙ � = �√� 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
122. If �, �, � > � and � ∈ ℕ, � ≥ � then prove: 
�
�
�(� + �)(� + �)
�
���
≤ � �
�
�(�� + �� + ��)
�
 
Proposed by Marin Chirciu-Romania 
Solution 1 by Tran Hong-Dong Thap-Vietnam 
��� ��(�) = �
�
�
�
, � > �; � ≥ �, � ∈ ℕ 
��
� (�) =
�
�
�
�
�
�
�
�
��
, ��
��(�) =
� − �
��
�
�
�
�
�
�
��
< �, ∀� > �, � ≥ � 
�
�
�(� + �)(� + �)
�
���
= � �
�
(� + �)(� + �)
�
���
≤
������
� �
�
�
�
�
(� + �)(� + �)
���
�
= 
= � �
�(� + � + �)
�(� + �)(� + �)(� + �)
�
≤
(�)
 � �
�
�(�� + �� + ��)
�
 
(�) ⇔ �(�� + �� + ��)(� + � + �) ≤ �(� + �)(� + �)(� + �) 
⇔ �(� − �)� + �(� − �)� + �(� − �)� ≥ � true for �, �, � > �. 
Solution 2 by Nas Meister-Thailand 
By using AGM inequality, we get: (� + �)(� + �)(� + �) ≥ ���� 
�(� + �)(� + �)(� + �) ≥ ��(� + �)(� + �)(� + �) + ���� = 
= �(� + � + �)(�� + �� + ��) 
�
�(�� + �� + ��)
≥
�(� + � + �)
(� + �)(� + �)(� + �)
=
(� + �) + (� + �) + (� + �)
(� + �)(� + �)(� + �)
= 
= �
�
(� + �)(� + �)
���
 
����
�(�� + �� + ��)
≥ (� + � + �)��� �
�
(� + �)(� + �)
���
≥
������
��
�
�(� + �)(� + �)
�
���
�
�
 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Hence, ∑
�
�(���)(���)
���� ≤ � �
�
�(��������)
�
 
Solution 3 by Nas Meister-Thailand 
We’ll prove by induction; Base case � = � 
By using BCS inequality, we have: 
�
�
�(� + �)(� + �)
���
≤ �� �
�
(� + �)(� + �)
���
≤ ��
�
�(�� + �� + ��)
 
⇔ �
�
(� + �)(� + �)
���
≤
�
�(�� + �� + ��)
 
⇔ �
� + �
(� + �)(� + �)(� + �)
���
=
�(� + � + �)
(� + �)(� + �)(� + �)
≤
�
�(�� + �� + ��)
 
⇔ �(� + � + �)(�� + �� + ��) ≤ �(� + �)(� + �)(� + �) 
⇔ ���� < (� + �)(� + �)(� + �) true from AGM inequality 
Now, assume that the statement is true for any � = � then we’ll probe that � = � + � is 
also true. By assumption and using Power Mean, since 
�
���
< �/� we have: 
�
�
�
� �
�
(� + �)(� + �)
���
���
�
���
≤ �
�
�
� �
�
(� + �)(� + �)
�
���
�
�
≤ 
≤ ��
�
�(�� + �� + ��)
�
�
�
=
�
�(�� + �� + ��)
 
Hence, ∑
�
�(���)(���)
������ ≤ � �
�
�(��������)
���
. Implies � = � + � is true. 
Solution 4 by Sanong Huayrerai-Nakon Pathom-Thailand 
�
�
�(� + �)(� + �)
�
���
≤ � �
�
�(�� + �� + ��)
�
 
Iff �� + �
� + √� + �
�
+ �� + �
� ≤ � �
�(���)(���)(���)
�(��������)
�
 
 
www.ssmrmh.ro 
 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
����� ∙ �(� + � + �)
�
≤ � �
�(� + �)(� + �)(� + �)
�(�� + �� + ��)
�
 
���� ∙ �(� + � + �) ≤ �� ∙
�(� + �)(� + �)(� + �)
�(�� + �� + ��)
 
�(� + � + �)(�� + �� + ��) ≤ �(� + �)(� + �)(� + �) 
�(��� + ��� + ��� + ��� + ��� + ��� + ����) ≤ 
≤ �(��� + ��� + ��� + ��� + ��� + ��� + ����) ����. 
 
123. If �, �, � > � then: 
��
�
+
��
�
+
��
�
≥ �
�� + ��
�
�
+ �
�� + ��
�
�
+ �
�� + ��
�
�
 
Proposed by Rahim Shahbazov-Baku-Azerbaijan 
Solution 1 by Abdul Hannan-Tezpur-India 
��
��
�
� ��
��
�
� �� ����� ≥
������
�� ���
�
= 
= ���(�� + ���� + �����
�
≥
���
��
∑ ��
�
� ���� ⇒ �
��
�
≥ ��
∑ ��
�
�
 
Now, let � = �
�����
�
�
, � = �
�����
�
�
, � = �
�����
�
�
 
Then by Power-Means inequality (PM), we obtain: 
��
∑ ��
�
�
= ��
∑ ��
�
�
≥
��
� ∙
∑ �
�
= � � 
Combining the two results, we obtain the desired inequality: 
��
�
+
��
�
+
��
�
≥ �
�� + ��
�
�
+ �
�� + ��
�
�
+ �
�� + ��
�
�
 
Solution 2 by Tran Hong-Dong Thap-Vietnam 
 
www.ssmrmh.ro 
 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
� ��
�� + ��
�
�
−
� + �
�
� = �
�� + ��
� − �
� + �
� �
�
��
�� + ��
�
�
+
� + �
� � �
��
� + ��
� + �
� + �
� �
�
�
= 
= �
�(�� + ��) − (� + �)�
�� ��
�� + ��
�
�
+
� + �
� � �
��
� + ��
� + �
� + �
� �
�
�
= 
= �
(��� + ���� + ���)(� − �)�
�� ��
�� + ��
�
�
+
� + �
� � �
��
� + ��
� + �
� + �
� �
�
�
= 
�
��
�
− (� + � + �) = �
(� − �)�
�
 
So, ∑
��
�
≥ ∑ �
�����
�
�
 
�
⎝
⎜
⎛�
�
−
��� + ���� + ���
�� ��
�� + ��
�
�
+
� + �
� � �
��
� + ��
� +
� + �
� �⎠
⎟
⎞
(� − �)� ≥ �; (�) 
Let �� =
�
�
−
������������
��(���)(�����)
; where � = �
�����
�
�
, � =
���
�
⇒ � ≥ � > � 
�� =
��(� + �)(�� + ��) − (��� + ���� + ���)
���(� + �)(�� + ��)
≥
���
 
≥
��(� + �) �
(� + �)�
� � −
(��� + ���� + ���)
���(� + �)(�� + ��)
= 
=
�� + ��� + ����� + �����
���(� + �)(�� + ��)
> �, ∀�, �, �, � > � 
Similarly: ��, �� > � ⇒ (�) is true. Proved. 
124. If �, �, � > 0 then: 
��
�
+
��
�
+
��
�
≥ ��(�� + �� + ��) − �(�� + �� + ��) 
Proposed by Rahim Shahbazov-Baku-Azerbaijan 
 
www.ssmrmh.ro 
 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Solution by Abdul Hannan-Tezpur-India 
Let � = � + � + �; � = �� + �� + �� 
The desired inequality can be rewritten as: 
�
��
�
≥ ���� − ��� ⇔ � �
��
�
− �� + �� ≥ ���� − ��� − � 
�
(� − �)�
�
≥
��� − ��� − ��
���� − ��� + �
=
�(�� − ��)
���� − ��� + �
= �
�(� − �)�
���� − ��� + �
 
⇔ � �
�
�
−
�
���� − ��� + �
� (� − �)� ≥ � 
So, it is enough to prove that �� ≔
�
�
−
�
����������
> � and analogs. 
�� > 0 ⇔ ���
� − ��� + � − �� ≥ � ⇔ ���� − ��� > 3� − � − � 
If �� − � − � ≤ � then ���� − ��� > 0 ≥ 3� − � − � 
If �� − � − � > 0 then ���� − ��� ≥ �� − � − � 
⇔ ��� − ��� ≥ (�� − � − �)� 
⇔ �(� + � + �)� − ��(�� + �� + ��) ≥ (�� − � − �)� ⇔ �(�� − �� + ��) ≥ � which is 
true. Similarly, ��; �� > 0. This proves the desired inequality. 
125. If�, �, � > � then: 
(��� + ���)(��� + ���)(��� + ���)(���� + ���� + ����)�
������(�� + ��)(�� + ��)(�� + ��)
≥ �� 
Proposed by Daniel Sitaru-Romania 
Solution 1 by Ertan Yildirim-Turkey 
Using Cebyshev’s inequality, we have: 
�(���� + ����) ≥ (�� + ��)(�� + ��). Hence, 
�(��� + ���) ≥ (�� + ��)(�� + ��) 
�(��� + ���) ≥ (�� + ��)(�� + ��) 
�(��� + ���) ≥ (�� + ��)(�� + ��) 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Then 
���������������������������
(�����)�������(�����)
≥
���������������������
�
 
(��� + ���)(��� + ���)(��� + ���)(���� + ���� + ����)�
������(�� + ��)(�� + ��)(�� + ��)
≥ 
≥
(�� + ��)(�� + ��)(�� + ��)(���� + ���� + ����)�
�������
 
≥
��� �√���� ∙ �√���� ∙ √������√���� ∙ ���� ∙ ����
�
�
�
�������
≥
������� ∙ ��������
�������
= �� 
Solution 2 by Sanong Huayrerai-Nakon Pathom-Thailand 
(��� + ���)(��� + ���)(��� + ���)(���� + ���� + ����)� ≥ 
≥
��(�� + ��)(�� + ��)(�� + ��)(�� + ��)(�� + ��)(�� + ��) ��(���)�
�
�
�
�
= 
=
��(�� + ��)(�� + ��)(�� + ��)(�� + ��)(�� + ��)(�� + ��)(���)�
�
≥ 
≥
���
�� ∙
�
�
(�� + ��)(�� + ��)(�� + ��)(���)��(���)� = 
= ��(���)�(�� + ��)(�� + ��)(�� + ��) 
 Therefore, 
�������������������������������������������
�
������(�����)�������(�����)
≥ �� 
 
126. If �, �, � > � then: 
��
�
+
��
�
+
��
�
≥
�� + ��
�� + ��
+
�� + ��
�� + ��
+
�� + ��
�� + ��
 
Proposed by Rahim Shahbazov-Baku-Azerbaijan 
Solution by Tran Hong-Dong Thap-Vietnam 
�� + ��
�� + ��
−
� + �
�
=
�(�� + ��) − (�� + ��)(� + �)
�(�� + ��)
=
(� − �)�(�� + �� + ��)
�(�� + ��)
 
� �
�� + ��
�� + ��
−
� + �
�
�
���
≥ �
(� − �)�(�� + �� + ��)
�(�� + ��)
���
 
 
www.ssmrmh.ro 
 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
So, ∑
��
����
≥ ∑
�����
��������
⇔ ∑ �
�
�
−
��������
��������
���� (� − �)
� ≥ � 
Let �� =
�
�
−
��������
��������
; �� =
�
�
−
��������
��������
; �� =
�
�
−
��������
��������
 
We prove that: �� > 0 (similarly ��, �� > 0) 
�� =
�
�
−
�� + �� + ��
�(�� + ��)
=
�(�� + ��) − (��� + ��� + ��)
��(�� + ��)
= 
=
[�� + �� − ��(� + �)] + ��
��(�� + ��)
≥
��������(���) ��
��(�� + ��)
> 0; ∀�, � > 0 
⇒ �� > �. Equality when � = � = � 
 
127. If �, �, � > � then: 
�(�� + �)
���� + �
+
�(�� + �)
���� + �
+
�(�� + �)
���� + �
≤ �(�� + �� + ��) �
�
��
+
�
��
+
�
��
� 
Proposed by Daniel Sitaru-Romania 
Solution by Tran Hong-Dong Thap-Vietnam 
��� =
�(�� + �)
���� + �
+
�(�� + �)
���� + �
+
�(�� + �)
���� + �
 
≤⏞
���
�(�� + �� + ��) ��
(�� + �)
���� + �
�
�
+ �
(�� + �)
���� + �
�
�
+ �
(�� + �)
���� + �
�
�
� 
= �(�� + �� + ��) ��
(�� + �)
���� + �
�
�
+ �
(�� + �)
���� + �
�
�
+ �
(�� + �)
���� + �
�
�
� 
Because: ��� = �(�� + �� + ��) �
�
��
+
�
��
+
�
��
� 
So, for � > �, we need to prove: �
(����)
������
�
�
≤
�
��
; (�������) 
↔
(�� + �)
���� + �
≤
�
�
; ↔ ��� + �� ≤ ���� + �; 
↔ ���� − ��� − �� + � ≥ �; ↔ (�� + �)(�� − �)� ≥ �; 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Which is clearly true by: � > � → �� + � > �, (�� − �)� ≥ �; 
Hence, ∑ �
(����)
������
�
�
≤ ∑
�
��
 
→
�(�� + �)
���� + �
+
�(�� + �)
���� + �
+
�(�� + �)
���� + �
≤ �(�� + �� + ��) �
�
��
+
�
��
+
�
��
� 
Proved. Equality if and only if � = � = � =
�
�
 
128. If �, �, � > � then: 
�(��� + ���� + ���)
���
≥ ������(� + �)(� + �)(� + �) 
Proposed by Rahim Shahbazov-Baku-Azerbaijan 
Solution 1 by Sanong Huayrerai-Nakon Pathom-Thailand 
For �, �, � > � we give � =
(���)(���)(���)
���
 and � = ��, � ≥ � 
�(��� + ���� + ���)
���
= �(�(� + �)� + ���)
���
≥ 
≥ ���(� + �)(� + �)(� + �)�
�
� + �(���)
�
��
�
≥ ������(� + �)(� + �)(� + �) 
��(� + �)(� + �)(� + �)�
�
� + �(���)
�
� ≥ �������(� + �)(� + �)(� + �)�
�
� 
� �
(� + �)(� + �)(� + �)
���
�
�
�
+ � �
���
(� + �)(� + �)(� + �)
�
�
�
≥ � 
�√�
�
+
�
√�
� ≥ � ⇔ �� +
�
�
≥ � ⇔ ��� − �� + � ≥ � true because � ≥ � 
 
Solution 2 by Florică Anastase-Romania 
�(��� + ���� + ���)
���
≥ ������(� + �)(� + �)(� + �) ⇔ 
�
�(� + �)� + ���
�(� + �)
���
≥ ���; (�) 
 
www.ssmrmh.ro 
 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Applying Huygens inequality, we get: 
�
�(� + �)� + ���
�(� + �)
���
= � �� ∙
� + �
�
+ � ∙
�
� + �
�
���
≥ 
≥ ���� ∙
(� + �)(� + �)(� + �)
���
�
+ ��� ∙
���
(� + �)(� + �)(� + �)
�
�
�
≥
(�)
��� 
(�) ⇔ ��
(� + �)(� + �)(� + �)
���
�
+ ��
���
(� + �)(� + �)(� + �)
�
≥ � 
Let � = �
(���)(���)(���)
���
�
≥
���
√�
�
= �, hence 
�� +
�
�
≥ � ⇔ ��� − �� + � ≥ � ⇔ (�� − �)(� − �) ≥ � true because � ≥ � 
From (1),(2) we obtain the desired inequality. 
 
129. If �, �, � > � then: 
��
�
+
��
�
+
��
�
≥ ���� − ��� + ��� + ���� − ��� + ��� + ���� − ��� + ��� 
Proposed by Rahim Shahbazov-Baku-Azerbaijan 
Solution by Abdul Hannan-Tezpur-India 
�
��
�
≥ � ���� − ��� + ��� ⇔ 
� �
��
�
− �� + �� ≥ � ����� − ��� + ��� −
� + �
�
� ⇔ 
�
(� − �)�
�
≥ �
��� − ��� + ��� − �
� + �
� �
�
���� − ��� + ��� +
� + �
�
= �
�(� − �)�
����� − ��� + ��� + �(� + �)
 
⇔ � �
�
�
−
�
����� − ��� + ��� + �(� + �)
� (� − �)� ≥ � 
So, it is enough to prove that: 
 
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�� ≔
�
�
−
�
���������������(���)
> � and analogs. 
�� > � ⇔ ����
� − ��� + ��� + �(� + �) − �� > � ⇔ 
����� − ��� + ��� > �� − �� 
 If �� − �� ≤ � then ����� − ��� + ��� > � > �� − �� 
 If �� − �� > � then ����� − ��� + ��� > �� − �� 
⇔ ��(��� − ��� + ���) ≥ (�� − ��)� = ���� − ���� + ��� 
⇔ ���� − ���� + ��� ≥ � ⇔ �(�� − �)� ≥ � which is true. 
Similarly, ��, �� > �. This proves the desired inequality. 
 
130. If �, �, � > � such that �� + �� + �� = ��� and � ≥ � then 
��
����
+
��
����
+
��
����
≥
�
��
 
Proposed by Marin Chirciu-Romania 
Solution by Nas Meister-Thailand 
Let � =
�
�
, � =
�
�
, � =
�
�
⇒ � + � + � = � 
��
����
+
��
����
+
��
����
≥
�
��
 
⇔ �
��
����
= �
�
�
�
��
��
=
�
���
� � ∙ �
�
�
�� 
By Weighted AM-GM inequality, we have: 
�
���
� � ∙ �
�
�
�� 
���
≥
�
���
� �����
���
= (������)� 
Let �(�) = ����� ⇒ ���(�) =
�
�
> �; ∀� ∈ ℝ� ⇒ � −convex function, by Jensen 
Inequality, we have: 
����� + ����� + �����
�
≥ �
� + � + �
�
���
� + � + �
�
� =
�
�
���
�
�
 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
����� + ����� + ����� ≥ ���
�
�
 
������ ≥
�
�
 
Therefore, ∑
��
����
= ∑
�
�
�
��
��
=
�
���
∑ � ∙ �
�
�
�� ≥ (������)� ≥
�
��
 
 
131. If �, �, � > � then: 
(��� + ��� + ���)(��� + ��� + ���)(��� + ��� + ���) 
(� + �)(� + �)(� + �)�(�� + ��)(�� + ��)(�� + ��)
≥ ��√� 
Proposed by Daniel Sitaru-Romania 
Solution 1 by Ertan Yildirim-Turkey 
��� + ��� + ��� = (� + �)� + �(�� + ��) ≥ ��(� + �)��(�� + ��) = 
= �√�(� + �)��� + �� 
�����������
(���)������
≥ �√�; Similarly: 
��� + ��� + ���
(� + �)√�� + ��
≥ �√�; 
��� + ��� + ���
(� + �)√�� + ��
≥ �√� 
Therefore, 
(��� + ��� + ���)(��� + ��� + ���)(��� + ��� + ���) 
(� + �)(� + �)(� + �)�(�� + ��)(�� + ��)(�� + ��)
≥ �√� ∙ �√� ∙ �√� = ��√� 
Solution 2 by Eldeniz Hesenov-Azerbaijan 
��� + ��� + ��� = (� + �)� + �(�� + ��) ≥ ��(� + �)��(�� + ��) = 
= �√�(� + �)√�� + ��. Hence, 
��� ≥
� ∙ �√� ∏(� + �) ∏ √�� + ��
∏(� + �) ∏ √�� + ��
 
Solution 3 by Sanong Huayrerai-Nakon Pathom-Thailand 
(��� + ��� + ���)(��� + ��� + ���)(��� + ��� + ���) = 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
= �(� + �)� + ��� + ��� + ��� + ���� �(� + �)� + ��� + ��� + ��� + ���� �(� + �)� + ��� + ��� + ��� + ���� ≥ 
≥ ��(� + �)�(� + �)�(� + �)�
�
+ ��(�� + ��)(�� + ��)(�� + ��)
�
�
�
≥ 
≥ ��√� ∙ �(� + �)�(� + �)�(� + �)�(�� + ��)(�� + ��)(�� + ��)
�
�
�
= 
= ��√�(� + �)(� + �)(� + �)�(�� + ��)(�� + ��)(�� + ��) 
Therefore, 
(��� + ��� + ���)(��� + ��� + ���)(��� + ��� + ���) 
(� + �)(� + �)(� + �)�(�� + ��)(�� + ��)(�� + ��)
≥ ��√� 
 
132. If �, �, � are non-negative real numbers such that � + � + � = ��, 
� ≥ � then: (�� +�� + ��)��� + �� + ���(�� + �� + ��) ≥ ����� 
Proposed by Dorin Mărghidanu-Romania 
Solution 1 by Florică Anastase-Romania 
(�� + �� + ��)��� + �� + ���(�� + �� + ��) ≥
�������
�������� +
�
�������
�
+ �������
�
�
�
 
= ��
�����
� + �
�����
� + �
�����
� �
�
= (�� + �� + ��)� ≥
������
 
≥ �
(� + � + �)�
���� ∙ �
�
�
= ����� 
Equality holds when � = � = �. 
 Solution 2 by Sanong Huayrerai-Nakon Pathom-Thailand 
(�� + �� + ��)��� + �� + ���(�� + �� + ��) ≥ ��
�����
� + �
�����
� + �
�����
� �
�
≥ ����� 
Iff �
�����
� + �
�����
� + �
�����
� ≥ ��� 
Iff �� + �� + �� ≥ ��� 
(� + � + �)�
����
≥ ��� 
(� + � + �)� ≥ (��)� 
Iff � + � + � ≥ �� true. 
 
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Equality holds when � = � = �. 
 Solution 3 by proposer 
By Holder inequality, we have: 
(�� + �� + ��)
�
���� + �� + ���
�
�(�� + �� + ��)
�
� ≥ 
≥ (������)
�
�+��������
�
�+(������)
�
� = �
�����
� + �
�����
� + �
�����
� = 
= �� + �� + ��; (�) 
Hence, 
(�� + �� + ��)��� + �� + ���(�� + �� + ��) ≥ (�� + �� + ��)�; (�) 
On the other hand, by the power-means inequality, for � ≥ � we have 
�
�� + �� + ��
�
�
�
�
>
� + � + �
�
⇒ �� + �� + �� ≥
(� + � + �)�
����
; (�) 
From (2),(3) we have: 
(�� + �� + ��)��� + �� + ���(�� + �� + ��) ≥ �
(� + � + �)�
����
�
�
= 
= �
(��)�
����
�
�
= ����� 
Equality holds when � = � = �. 
133. If �, �, � > 0 then: 
(�� + �� + ��)���
(���)���
≥ � √�
���
+ √�
���
+ √�
���
�
���
 
Proposed by Daniel Sitaru-Romania 
Solution 1 by Oyebamiji Oluwsaeyi-Nigeria 
� √��
���
+ √��
���
+ √��
���
� ≥
���
� √��
���
∙ √��
���
+ √��
���
∙ √��
���
+ √��
���
∙ √��
���
� = 
= √���
���
� √�
���
+ √�
���
+ √�
���
�; (�) 
Also 
� √��
���
+ √��
���
+ √��
���
�
���
≤
���������
����(�� + �� + ��) 
 
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�� + �� + �� ≥
� √��
���
+ √��
���
+ √��
���
�
���
����
 
�� + �� + �� ≥
(�) � √���
���
� √�
���
+ √�
���
+ √�
���
��
���
����
 
(�� + �� + ��)��� ≥
(���)��� �� √�
���
+ √�
���
+ √�
���
�
���
�
���
����∙���
≥
���
 
≥
��� (���)
��� ���� √���
����
�
���
� √�
���
+ √�
���
+ √�
���
��
���
����∙���
 
=
(���)��� �����(���)
��
���� √�
���
+ √�
���
+ √�
���
��
���
����∙���
 
=
����∙���(���)���(���)��� √�
���
+ √�
���
+ √�
���
�
���
����∙���
 
= (���)���� √�
���
+ √�
���
+ √�
���
�
���
 
Therefore, 
(��������)���
(���)���
≥ � √�
���
+ √�
���
+ √�
���
�
���
 
Solution 2 by Sanong Huayrerai-Nakon Pathom-Thailand 
For �, �, � > 0 we give: � = ����, � = ����, � = ���� 
Hence, 
(��������)���
(���)���
≥ � √�
���
+ √�
���
+ √�
���
�
���
 
�(��)��� + (��)��� + (��)����
���
((���)���)���
≥ (� + � + �)��� 
(��)��� + (��)��� + (��)���
(���)���
≥ � + � + � 
(��)���
����
+
(��)���
����
+
(��)���
���� 
≥ � + � + � 
((��)��� + (��)��� + (��)���)�
(��)������� + (��)������� + (��)�������
≥ � + � + � 
(��)��� + (��)��� + (��)��� + �(������������ + ������������ + ������������) ≥ 
≥ (� + � + �)((��)������� + (��)������� + (��)�������) 
 
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(��)��� + (��)��� + (��)��� + ������������ + ������������ + ������������ ≥ 
≥ ������������ + ������������ + ������������ + ������������ + ������������ + ������������ 
True, because 
�(���� + ���� + ����) ≥ (����� + ����� + �����) + (����� + ����� + �����) 
Solution 3 by Michael Sterghiou-Greece 
(�� + �� + ��)���
(���)���
≥ � √�
���
+ √�
���
+ √�
���
�
���
; (�) 
Lemma: If �� > 0, � = �, ������ and ���� ∙ … ∙ �� = � then � ≥ � 
� ��
�
�
���
≥ � ��
�
�
���
 
For the proof �� +
���
�
≥
�
�
∙ ��
� by weighted AGM (
���
�
−times the 1) which by summation 
on � to � we get: 
�� ��
�
�
���
� − �� ��
�
�
���
� ≥ �
�
�
− �� �� ��
�
�
���
− �� ≥ � 
� ��
�
�
���
≥ � �� ��
�
�
���
�
�
�
= � 
Now (1) homogeneous as assume ��� = �, (�) ⇒ ∑ ����� ≥ ∑ √�
���
��� 
But �∑ ����� �
�
≥ �(� + � + �); (��� = �); 
� + � + � ≥
�
�
�� √�
���
�
�
; (�(�) = �� − ������ ��� � √�
���
≥ � √�
���
���
 
From lemma as 
�
�
≥
�
���
 the result is (1). Done! 
134. If �, �, � > 0 then prove: 
� �
�� + ��
�� + ��
�
���
≥
�
�
�� +
�� + �� + ��
�� + �� + ��
� 
Proposed by Marin Chirciu-Romania 
 
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Solution by Abdul Hannan-Tezpur-India 
� �
�� + ��
�� + ��
�
���
= �
�� + ��
�(�� + ��)(�� + ��)(�� + ��)
�
���
≥
���
 
�
�� + ��
�� + �� + �� + �� + �� + ��
����
= �
�(�� + ��)
��� + (� + �)�
���
≥
���
 
�
�(�� + ��)
��� + �(�� + ��)
���
=
�
�
�
�� + ��
�� + �� + ��
���
= 
=
�
�
∙
�� + �� + �� + �� + �� + ��
�� + �� + ��
=
�
�
�� +
�� + �� + ��
�� + �� + ��
� 
 
135. If �, �, � > �, ��� = � then prove: 
(� + � + �)�
�� + �� + ��
≥ �� 
Proposed by Daniel Sitaru-Romania 
Solution 1 by Israel Castillo Pico-Brazil 
(� + � + �)� ≥ �����(�� + �� + ��) 
� �� + � � ��� + � � ��� + �� � ���� + �� � ���� + �� � ���� + �� � ����� ≥ �� � ����� 
In terms of Murihead’s theorem 
[�, �, �] ≥ [�, �, �]; ��[�, �, �] ≥ ��[�, �, �] 
��[�, �, �] ≥ ��[�, �, �]; ��[�, �, �] = ��[�, �, �]; 
��[�, �, �] ≥ [�, �, �] 
Solution 2 by Sudhir Jha-Kolkata-India 
(� + � + �)�
�� + �� + ��
=
(� + � + �)�(� + � + �)�
�� + �� + ��
= 
=
�� + �� + �� + ��� + ��� = ���
�� + �� + ��
∙ (� + � + �)� = 
 
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= �
�� + �� + ��
�� + �� + ��
+ �� (� + � + �)� ≥ (� + �) ∙ ����� = �� 
Where, 
��������
��������
≥ � and (� + � + �)� ≥ ����� by AGM inequality. 
Equality holds when � = � = � = �. 
 Solution 3 by Abdul Aziz-Semarang-Indonezia 
(� + � + �)�
�� + �� + ��
=
(� + � + �)�(� + � + �)�
�� + �� + ��
≥
���
 
≥
������
� �
�
�(�� + �� + ��)
�� + �� + ��
= ��. 
Equality holds when � = � = � = �. 
 Solution 4 by Sanong Huayrerai-Nakon Pathom-Thailand 
For ��� = �; �, �, � > 0 we will show that 
�(� + � + �) ≥ �� + �� + �� + �� 
Give � =
�
�
, � =
�
�
, � =
�
�
; �, �, � > 0 hence, 
� �
�
�
+
�
�
+
�
�
� ≥ �� + �
�
�
+
�
�
+
�
�
� 
�(��� + ��� + ���) ≥ ����� + ��� + ��� + ��� 
We give � ≤ � ≤ � and � ≤ ��, � ≤ ���; �, � ≥ � hence, 
�(��(���) + (���)�(��) + (��)��) ≥ 
≥ ���(��)(���) + ��(��) + (��)�(���) + (���)�� 
�(� + ���� + �) ≥ ���� + � + ��� + ��� 
� + � + ���� + �(� + � + ����) ≥ ���� + � + ��� + ��� 
���� − � ≥ (�� − �)(� + �) 
(�� − �)(�� + �) ≥ (� + �)(�� − �) 
�� + � ≥ � + � 
Consider 
(�����)�
��������
≥ �� ⇔
�����
��������
∙ �
�����
�
�
�
≥ � 
 
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⎝
⎜⎜
⎛ � + �
�
� + � + �
�� + �� + �� +
�
� + � + �
� ⎠
⎟⎟
⎞
�
≥ � 
�
�� + �� + ��
� + � + � +
��
� + � + �
≥ � 
⇔ �(� + � + �) ≥ �� + �� + �� + �� true. 
 
136. If ��, �� ∈ (�, ∞), ∑ ��
�
��� = � and � ∈ ��; (� −is a permutation of the 
set {�, �, … , �}), then holds the inequalities: 
�
�
∙ � ��
�
���
≥
�
�!
∙ � � �
�
��(�)
�
����∈��
≥ �� ��
�
���
�
≥
�!
∑
�
∏ �
�
��(�)�
���
�∈��
≥
�
∑
�
��
�
���
 
Proposed by Dorin Marghidanu-Romania 
Solution by proposer 
For the first inequality in (�), by using means inequality-in weighted from 
� ��
��
�
���
≤ � ����
�
���
; (∗) 
We have: 
� � �
�
��(�)
�
����∈��
≤
(∗)
� � ��(�) ∙ ��
�
����∈��
≤ � � ��(�) ∙ ��
�∈��
�
���
= 
(�� ������ ��, � = �, ������, ��� ������� ������� ��� �� �������� ��(� − �)! �����) 
= ���(�) + ��(�) + ⋯ + ��(��)� ∙ �� ∙ (� − �)! + ⋯ + ���(�) + ��(�) + ⋯ + ��(��)� ∙ �� ∙ (� − �)! 
= (� − �)! ∙ (�� + �� + ⋯ + ��) 
Dividing by �!, on obtain the first inequality. 
Now, using means inequality for �! = ����(��) numbers, we have: 
 
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� � �
�
��(�)
�
����∈��
≥ �! ∙ �� � ��
��(�)
�
����∈��
�!
= 
(�� ������ ��, � = �, ������, ��� ������� ������� ��� �� �������� ��(� − �)! �����) 
= �! ∙ ���
(���)!
∙ ��
(���)!
∙ … ∙ ��
(���)!�!
= �! ∙ ��� ��
����
�
(���)!
�!
= �! ∙ �� ��
�
���
�
 
Therefore, the second inequality from enounce. 
(�) For the another two inequalities from enounce, on make the substitution 
�� →
�
��
 in the two inequalities in (�) 
137. If � ∈ ℕ, � ≥ � then: 
(� − �)
�
�
�
�
��� > �
�
���
�
�
��� 
Proposed by Daniel Sitaru-Romania 
Solution 1 by Adrian Popa-Romania 
Let be the function: �(�) = �
�
���
�
�
�, � ≥ � 
�(�)�
�
�
���
�
�
�
�����
⇒ ��(�) = �−
����
(� + �)�
−
����
��
+
�
�(� + �)
+
�
��
� ∙ �(�) 
�
��
−
����
��
=
�
��
(� − ����) < �, ∀� ≥ � 
�
�(� + �)
−
����
(� + �)�
<
�
��
−
����
��
< � 
Hence, ��(�) < �, ∀� ≥ � ⇒ � ↘, ∀� ≥ � ⇒ �(� − �) > �(�) > �(� + �) ⇒ 
(� − �)
�
�
�
�
��� > �
�
���
�
�
� > �
�
���
�
�
���, � ∈ ℕ, � ≥ � 
Solution 2 by Nikos Ntorvas-Greece 
Let be �(�) = �
�
���
+
�
�
� ����, � > � 
��(�) =
�
��
(� − ����) +
�
� + �
�
�
�
−
����
� + �
� ; (�) 
 
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Let be �(�) =
�
�
−
����
���
, � > � 
��(�) = −
�
��
+
����
(� + �)�
−
�
�(� + �)
=
������ − (� + �)� − �
(�� + �)�
 
So, we have that ��(�) < � because ���� ≤ � − � < � + �, ∀� > � ⇒ � ↘ (�, ∞) 
For � ≥ � we have that: 
� ≥ � ⇒ ���� ≥ ���� ⇒ � − ���� ≤ � − ���� < �; (� < �) 
�
��
(� − ����) < � 
� ≥ � ⇒ �(�) ≤ �(�) ⇔ �(�) ≤
�
�
−
�
�
���� ⇔ �(�) ≤
� − �����
��
< � ⇔ 
�
� + �
�
�
�
−
����
� + �
� < � 
Finally, � −is strictly decreasing for � ≥ �. 
� − � < � < � + �; (� ↘) ⇒ �(� − �) > �(�) > �(� + �) 
�
�
�
+
�
� − �
� ���(� − �) > �
�
�
+
�
� − �
� ���� > �
�
� + �
+
�
� + �
� ���(� + �) 
���(� − �)
�
�
�
�
��� > ����
�
�
�
�
��� > ���(� + �)
�
���
�
�
��� 
(� − �)
�
�
�
�
��� > �
�
�
�
�
��� > (� + �)
�
���
�
�
��� 
 
138. If �, �, � > � then prove: 
�(� + � + �)� ≥ �����{��� + ��� + ��� + ���, ��� + ��� + ��� + ���} 
Proposed by Nguyen Van Canh-Ben Tre-Vietnam 
Solution by Rahim Shahbazov-Baku-Azerbaijan 
�(� + � + �)� ≥ �����{��� + ��� + ��� + ���, ��� + ��� + ��� + ���}; (�) 
 
Denote � = ��� + ��� + ��� + ���, � = ��� + ��� + ��� + ��� 
WLOG suppose that: � ≥ � ≥ � then � ≥ � because 
� − � = (� − �)(� − �)(� − �) ≥ � 
We must show that 
 
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�(� + � + �)� ≥ ��(��� + ��� + ��� + ���); (�) 
� ≥ � ≥ � ⇒ (� − �)(� − �) ≥ � ⇒ �� + �� ≤ �� + ��
(�)
�� 
��� + ��� + ��� + ��� = ��� + �(�� + ��) + ��� ≤ 
≤ ��� + �(�� + ��) + ��� = �(� + �)� = 
=
�
�
∙ ��(� + �)(� + �) ≤
�
�
∙
�
��
(�� + � + � + � + �)� 
Therefore, �(� + � + �)� ≥ ��(��� + ��� + ��� + ���) 
 
139. If � < � ≤ � then: 
(� − �)�
���
+ ��� �
�
�
� ≤
� − �
�
 
Proposed by Daniel Sitaru-Romania 
Solution 1 by Vivek Kumar-India 
Inequality can be rewritten as: 
�
�
�� −
�
�
�
�
− ���
�
�
≤
�
�
− �; (�) 
Let 
�
�
= � such that � < � ≤ �; (�) 
(�) ⇒
�
�
(� − �)� − ���� ≤
�
�
− � ⇔ ���� −
(� − �)�
�
+
� − �
�
≥ �; (�) 
Let �(�) = ���� −
(���)�
�
+
���
�
, ��(�) = −
(���)(���)�
��
< � ⇒ � − decreasing, hence 
∀� ≤ � ⇒ �(�) ≥ �(�) ⇒ ���� −
(� − �)�
�
+
� − �
�
≥ � 
Solution 2 by Adrian Popa-Romania 
(� − �)�
���
+ ��� �
�
�
� ≤
� − �
�
 
�� �
�
� − ��
�
���
+ ���
�
�
≤
�
�
− � 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
�
�
� − ��
�
���
��
+ ���
�
�
≤
�
�
− � 
Let �(�) =
(���)�
���
+ ���� − � + � 
��(�) = −
(� − �)�(� + �)
��
≤ �, ∀� ≥ � 
For � =
�
�
⇒
�
�
�
���
�
��
�
�
�
� + ���
�
�
≤
�
�
− � 
Solution 3 by Rovsen Pirguliyev-Sumgait-Azerbaijan 
���� ≤
����
��
; (�) 
for � =
�
�
≥ � we have: ��� �
�
�
� ≤
�
�
�
�
�
��
��
�
=
�����
���
; (�) 
Hence, 
(� − �)�
���
+ ��� �
�
�
� ≤
(� − �)�
���
+
�� − ��
���
; (�) 
We want to prove that: 
(� − �)�
���
+
�� − ��
���
≤
� − �
�
 
�(� − �)� + �(�� − ��) ≤ ���(� − �) 
(� − �)� ≤ � (����) ∀� ≤ � 
Solution 4 by Khaled Abd Imouti-Damascus-Syria 
�� − �� + � ≥ � ⇒ �� + � ≥ �� ⇒
�� + �
���
≥
��
���
⇒
�
�
+
�
��� 
≥
�
�
 
�
�
≤
�
�
+
�
���
⇒ �
��
�
�
�
≤ � �
�
�
+
�
���
� ��
�
�
 
���� ≤
����
��
 for � =
�
�
≥ � we have: 
 
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��� �
�
�
� ≤
�
�
��
�
− �
��
�
⇔ ��� �
�
�
� ≤
�� − ��
���
 
(� − �)�
���
+ ��� �
�
�
� ≤
(� − �)�
���
+
�� − ��
���
 
We prove that: 
(� − �)�
���
+
�� − ��
���
≤
� − �
�
⇔ �(� − �) + �(� + �) ≤ ��� ⇔ 
�� + �� ≤ ��� ⇔ � ≤ � (����) 
So, 
(���)�
���
+ ��� �
�
�
� ≤
���
�
 
 Solution 5 by Nikos Ntorvas-Greece 
For � = � the desired inequality is obviously. Let � < �. 
(� − �)�
���
+ ��� �
�
�
� ≤
� − �
�
 
⇔
�
�
�� −
�
�
�
�
+ ���� − ���� −
�
�
+ � ≤ �; (�) 
Let be �(�) =
�
�
�� −
�
�
�
�
+ ���� − ���� −
�
�
+ �; � < � < � 
��(�) =
(���)�(���)
����
> � ⇒ � −increasing on (�, �) hence, 
� < � ⇒ �(�) < �(�) ⇔ �(�) < � ⇒ (�)���� ∀� ∈ (�, �) and for � = � we get: 
(� − �)�
���
+ ��� �
�
�
� ≤
� − �
�
 
140. If �, �, � ∈ ��,
�
�
� , ������������ = ������������ then: 
� + (� + �����)(� + �����)(� + �����) ≥
�
���������������
 
Proposed by Daniel Sitaru-Romania 
Solution by George Florin Şerban-Romania 
���� = �, ���� = �, ���� = �; �, �, � > 0, ��� = 1, �� + � =
�
�����
 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
� + (� + ��)(� + ��)(� + ��) ≥ �(�� + �)(�� + �)(�� + �) 
� + � ��
���
+ �(��)�
���
+ � ≥ � + � �(��)�
���
+ � � �� + �
���
 
� ��
���
+ �
�
��
���
+ � ≥ � � ��
���
+ � �
�
��
���
 
� ��� +
�
��
+ � −
�
��
− ����
���
≥ �; (�) 
�� +
�
��
+ � −
�
��
− ��� ≥ �, ∀� > 0 
�� − ��� + ��� − �� + � ≥ � ⇔ (� − �)� ���� −
�
�
�
�
+
�
�
� ≥ � 
True from (� − �)� ≥ � and ��� −
�
�
�
�
+
�
�
≥
�
�
> 0. Hence, 
� ��� +
�
��
+ � −
�
��
− ����
���
≥ � 
Therefore, � + (� + �����)(� + �����)(� + �����) ≥
�
���������������
 
 
141. If � < � ≤ � then prove: 
��
��
≥ ��√���
���
 
Proposed by Daniel Sitaru-Romania 
Solution 1 by Pavlos Trifon-Greece 
Let � =
�
�
∈ (�, �] ⇔ � = ��. 
��
��
≥ ��√���
���
⇔ ���
��
��
≥ �����√���
���
 
⇔ ����� − ����� ≥ (� − �) �� +
�
�
���(��)� 
����� − (��)���(��) ≥ (� − ��) �� +
���(��) + ����
�
� 
 
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����� − ������ − ������ ≥ �(� − �) �� + ���� +
����
�
� 
���� ≤
�� − �
� + �
, ∀� ∈ (�, �]; (�) 
Let �(�) = ���� −
����
���
, � ∈ (�, �] 
��(�) =
(� − �)�
�(� + �)�
> 0; ∀� ∈ (�, �] ⇒ � ↑ (�, �] 
Therefore, for � < � ≤ 1 we have �(�) ≤ �(�) ⇒ (�) is true. 
 Solution 2 by Nikos Ntorvas-Greece 
��
��
≥ ��√���
���
⇔
��
��
≥ ���� ∙
�
�
�
�
�
�
∙
�
�
�
�
�
�
 
⇔ �
�
�
�
���
�
≥ ���� ⇔
� + �
�
(���� − ����) ≥ � − � 
⇔
���� − ����
� − �
≥
�
� + �
 
⇔
���
���������
≤
���
�
 –Logarithm means inequality 
Therefore, 
��
��
≥ ��√���
���
 
 Solution 3 by Sudhir Jha-Kolkata-India 
��
��
≥ ��√���
���
; (�) 
⇔
���
���
�
���
���
�
≥ ���� ⇔ �
�
�
�
���
�
≥ ���� 
⇔ �
�
�
�
���
�
≥ �
��
���
�
�
⇔ �
�
�
�
��
�
�
≥ �
��
�
�
���
; (�) 
Now, let �(�) = ���� − ��(���), ∀� ≥ � 
��(�) = �������� + (� + �)�� − ������ ≥ �; ∀� ≥ � 
Hence, � −increasing for all � ≥ � ⇒ �(�) ≥ �(�) = �; ∀� ≥ � 
���� ≥ ��(���); ∀� ≥ � 
 
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�
�
�
�
��
�
�
≥ ��(
�
�
��);
�
�
≥ � ⇒ (�) is true ⇒ (�) is true. 
Equality holds when � = �. 
142. If �, � ∈ ℝ, �� + �� + ��� ≤ ��� + ��� then: 
�� + �� + ����� + ����� ≤ ���(��� + ��� − ���) 
Proposed by Daniel Sitaru-Romania 
Solution by Tran Hong-Dong Thap-Vietnam 
 
 
 
 
 
 
 
 
 
 
 
For �, � ∈ ℝ, we have: �� + �� + ��� ≤ ��� + ��� ↔ (� − �)� + (� − ��)� ≤ �� 
→ �� ��� = � ↔ � = ��; �� ��� = � ↔ �� = ��; 
�� ��� = � ↔ � = �; �� ��� = �� ↔ �� = �; Other, 
�� + �� ≤ ��� + ��� − ���; (1) 
 �� + �� + ����� + ����� ≤ ���(��� + ��� − ���); (�) 
From (1) & (2) we need to prove: 
�� + �� + ����� + ����� ≤ ���(�� + ��) ↔ (�� + �� − ���)(��+ �� − ��) ≤ � 
Which is clearly true because: 
 �� + �� ≥ �����
� + �� ���
� = �� + �� = �� > �� → �� + �� − �� > �; 
 �(�; �) ∈ {(�; �) ∈ ℝ�|(� − �)� + (� − ��)� ≤ ��} ⊊ 
 
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{(�; �) ∈ ℝ�|�� + �� ≤ ���}. Proved. 
143. Solve for integers: 
�� = �� + �� 
Proposed by Jalil Hajimir-Toronto-Canada 
Solution by Bedri Jajrizi-Mitrovica-Kosovo 
Let: � + � = � ⇒ �� = (� − �)(� + �); (� − �, � + �) ∈ {�, �} 
Case i) (� − �, � + �) = � ⇒ � −even. 
(� − �, � + �) = ���� ⇒ ∃�, � ∈ ℤ such that � + � = ��, � − � = �� ⇒ �� − �� = �. 
It’s clear that: �� = �, �� = −� ⇒ � = �, � = −� 
So: � + � = � ⇒ � = �, � = −� 
Solutions is: � = �, � = −� 
Case ii) (� − �, � + �) = � ⇒ � is odd, � is even. 
It’s clear that � ≡ �, �(����). 
Also (�, �) is solution⇔ (−�, �) is solution. 
So assume that � ≡ �(����) ⇒ � + � ≡ �(����), � − � ≡ �(����) 
We can write the equation in the form: 
�
�
�
�
�
=
���
�
∙
���
�
 where �
���
�
,
���
�
� = � because (� + �, � − �) = � 
Let: 
���
�
= ��,
���
�
= �� ⇒ ��� − � = � = ��� + � ⇒ �� − ��� = � 
(�, �) ∈ {(�, �), (−�, �)} ⇒ � ∈ {−�, −�} ⇒ � ∈ {−�, −�} 
Solutions are: (�, �) ∈ {(−�, �), (−�, �), (�, �), (�, �)} 
 
144. Solve for natural numbers: 
(� − �)‼ (� − �)‼
(� − �)‼ (� − �)‼
+
(� − �)‼ (� − �)‼
(� − �)‼ (� − �)‼
+
(� − �)‼ (� − �)‼
(� − �)‼ (� − �)‼
= �� 
Proposed by Daniel Sitaru-Romania 
Solution 1 by Bedri Hajrizi-Mitrovica-Kosovo 
 
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(� − �)‼ (� − �)‼
(� − �)‼ (� − �)‼
+
(� − �)‼ (� − �)‼
(� − �)‼ (� − �)‼
+
(� − �)‼ (� − �)‼
(� − �)‼ (� − �)‼
= �� ⇔ 
(� − �)(� − �) + (� − �)(� − �) + (� − �)(� − �) = �� ⇔ 
��� − ��� = � ⇔ ��(� − �) = �, � ∈ ℕ ⇔ � = �. 
 Solution 2 by Rovsen Pirguliev-Sumgait-Azerbaijan 
Since (� − �)‼ = … (� − �)������� ∙ (� − �) 
(� − �)‼ = … (� − �)������� ∙ (� − �) 
Similarly: (� − �)‼ = … (� − �)������� ∙ (� − �) 
(� − �)‼ = … (� − �)������� ∙ (� − �) we have 
(� − �)(� − �) + (� − �)(� − �) + (� − �)(� − �) = �� ⇔ 
��� − ��� = � ⇔ ��(� − �) = �, � ∈ ℕ ⇔ � = �. 
145. Solve for natural numbers: 
(�� − �) ∙ �
��� − ���� − ��� + ��
��� + ���� − ��� − ��
�
���
= ��� − �� 
Proposed by Costel Florea-Romania 
Solution 1 by Rovsen Pirguliyev-Sumgait-Azerbaijan 
Since 
��� − ���� − ��� + �� = (�� − �)(�� − �)(�� + �) 
��� + ���� − ��� − �� = (�� − �)(�� + �)(�� + �) 
We have: (�� − �) ∙ ∏
���������������
���������������
�
��� = ��� − �� 
(�� − �) ∙
� ∙ � ∙ �
� ∙ � ∙ ��
∙
� ∙ � ∙ ��
� ∙ � ∙ ��
∙ … ∙
(�� − �)(�� − �)(�� + �)
(�� − �)(�� − �)(�� + �)
∙
(�� − �)(�� − �)(�� + �)
(�� − �)(�� + �)(�� + �)
= ��� − �� 
Hence, (�� − �) ∙
�
(����)
= ��� − �� 
(�� + �)(��� − ��) = �� − � 
���� − ��� − ��� = � 
 
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No solution for natural numbers. 
Solution 2 by Izumi Ainsworth-Lima-Peru 
(�� − �) ∙ �
��� − ���� − ��� + ��
��� + ���� − ��� − ��
�
���
= ��� − �� 
(�� − �) ∙ �
(�� − �)(�� − �)(�� + �)
(�� − �)(�� + �)(�� + �)
�
���
= ��� − �� 
(�� − �) �
�� − �
�� + �
�
���
�
�� − �
�� + �
�
���
�
�� + �
�� − �
�
���
= ��� − �� 
(�� − �) ∙
�‼ (�� − �)‼
(�� + �)‼
∙
�
�� + �
∙
(�� + �)‼
�‼ (�� − �)‼
= ��� − �� 
�
(�� + �)(�� + �)
= �� − �� 
No solution for natural numbers. 
146. � = � � ∣∣ � ∈ ℝ, √� + �
�
+ √� − �
�
= √�
�
� 
� = � � ∣∣ � ∈ ℝ, √� + �
�
+ √� − �
�
= √�
�
� 
Find the sets Ω�, Ω� such that: 
�∆Ω� = � Ω�∆� = �, ��∆� = (� �⁄ ) ∪ (� �⁄ )� 
Proposed by Daniel Sitaru-Romania 
Solution 1 by Adrian Popa-Romania 
√� + �
�
+ √� − �
�
= √�
�
 
Let: √� + �
�
= �, √� − �
�
= � ⇒ �� + � = √�
�
�� + �� = �
⇒ �
(� + �)� = �
�� + �� = �
; (�) ⇒ 
���� + ���� + ������ + ������ + ������ + ������ = � 
������ + ��� + ������(�� + ��) + ������(� + �) = � 
��(� + �)(�(�� − ��� + ���� − ��� + ��) + ����(�� − �� + ��) + ������) = � 
If �� = � then � = � ⇒ � = √�
�
 and � = � ⇒ � = √�
�
 
� + � = √�
�
≠ � 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
�(�� − ��� + ���� − ��� + ��) + ����(�� − �� + ��) + ������ = � 
�� + �� − ��� − ��� + ���� + ���� − ����� + ���� + ����� = � 
�� + �� + ���� + ���� + ����� = � 
�� + �� + ����� + ���(�� + ��) + ���� = � 
(�� + �� + ��)� = � impossible, because ∆< 0. 
So, � √
� + �
�
= �
√� + �
�
= √�
� ⇒ � ∈ {−�, −�}. Now, √� + �
�
+ √� − �
�
= √�
�
 
Let: √� + �
�
= �; √� − �
�
= � ⇒ �� + � = √�
�
�� + �� = �
⇒ (� + �)� = � ⇒ 
���� + ���� + ������ + ������ + ������ + ������ + ������� + ������� = � 
�� ���� + ��� + ������� + ��� + ������(�� + ��) + ������(� + �)� = � 
��(� + �) �
���� − ��� + ���� − ���� + ���� − ��� + ���
+������� − ��� + ���� − ��� + ��� + ��������� − �� + ��� + ������
� = � 
If �� = � then � = � ⇒ � = √�
�
 and � = � ⇒ � = √�
�
 
� + � = √�
�
≠ � 
��� + ��� + ������ + ������ + ������ + ���� 
= � (�� + �� + ��)������������
��
+ ����(� + �)����������
��
= � ⇔ � = � = � 
So, √� + �
�
= � ⇔ � = −� or √� + �
�
= √�
�
⇔ � = � then � ∈ {�, �} 
� = {−�, �}, � = {−�, �} 
�∆Ω� = � ⇒ ({−�, �} ∕ Ω�) ∪ (Ω� {−�, �}⁄ ) = {−�, �} ⇒ Ω� = {−�, �, −�, �} 
Ω�∆� = �⇒({−�, �} Ω�⁄ ) ∪ (Ω� {−�, �}⁄ ) = {−�, �} ⇒ Ω� = {−�, �, �, �} 
Solution 2 by Ravi Prakash-New Delhi-India 
√� + �
�
+ √� − �
�
= √�
�
 
Put: √� + �
�
= �, √� − �
�
= � ⇒ � + � = ��, � − � = �� ⇒ �� + �� = � 
Also, � + � = √�
�
⇒ (� + �)� = � = �� + �� ⇒ 
�
�
�
� ��� + �
�
�
� ���� + �
�
�
� ���� + �
�
�
� ���� + �
�
�
� ���� + �
�
�
� ��� = � 
������� + ��� + ���(�� + ��) + �����(� + �)� = � 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
���(� + �)[�� − ��� + ���� − ��� + �� + ���� − ����� + ���� + �����] = � 
��[�� − ���� + ����� + ���� + ��] = � 
��(�� + �� + ��)� = � how �� + �� + �� > 0 ⇒ �� = 0 ⇒ � = 0 or � = � ⇒ 
� ∈ {−�, �} = �. Similarly, √� + �
�
+ √� − �
�
= √�
�
⇒ � ∈ {−�, �} = � 
�∆Ω� = � ⇒ ({−�, �} ∕ Ω�) ∪ (Ω� {−�, �}⁄ ) = {−�, �} ⇒ Ω� = {−�, �, −�, �} 
Ω�∆� = �⇒({−�, �} Ω�⁄ ) ∪ (Ω� {−�, �}⁄ ) = {−�, �} ⇒ Ω� = {−�, �, �, �} 
147. 
�(�) = �� − ��� + ���� + ���
� + ���
� + ���
� + ���
� + ��� + �� ∈ ℝ[�] 
If �(�) has all roots real prove that its lie in [−�, �]. 
Proposed by Rajeev Rastogi-India 
Solution by Khanh Hung Vu-Ho Chi Minh-Vietnam 
Support �(�) has all roots which is ��, � = �, ������ 
By Viete theorem for polynomial �(�), we have: 
⎩
⎪
⎨
⎪
⎧
� ��
�
���
= �; (�)
� ����
�,�∈[�,�]
���
= ��; (�)
 
We have (2) equivalent to: 
�� ��
�
���
�
�
− � ��
�
�
���
= � ∙ ��
(�)
�� � ��
�
�
���
− �� = �� 
On the other hand, by CBS inequality, we have: 
� ��
�
�
���
≥
�
�
�� ��
�
���
�
�
��� ����� � ��
�
���
= � − �� ⇒ � ��
�
�
���
≥ ��
� +
(� − ��)
�
�
 
��
� +
(� − ��)
�
�
≤ �� ⇒ �� ∈ �
� − √���
�
,
� + √���
�
� �� �� ∈ [−�, �] 
Similarly, for, ��, ��, … , �� ∈ [−�, �] 
 
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So, �� ∈ [−�, �], � = �, ������ 
148. Find all real roots: 
����(� − �) + ����(�� + � + �) − ��� + �� = � 
Proposed by Daniel Sitaru-Romania 
Solution by Abner Chinga Bazo-Lima-Peru 
����(� − �) + ����(�� + � + �) − ��� + �� = � 
���� − ���� + ���� + ���� + ���� − ��� + �� = � ; (������) ⇔ 
(�� − �)�(���� + ��� + ��� + ��) = � 
���� + (��� + ��� + �) + �� = � 
���� + (�� + �)� + �� > �, ∀� ∈ ℝ ⇒ �� − � = � ⇒ � =
�
�
 
So, � = �
�
�
� 
149. Solve for real numbers: 
|����| + |����| = �(� + ���� + ����)(� − ���� − ����) 
Proposed by Daniel Sitaru-Romania 
Solution 1 by Florentin Vişescu-Romania 
|����| + |����| = �(� + ���� + ����)(� − ���� − ����) ⇔ 
|����| + |����| = �� − ��������� − ����� − ����� ⇔ 
����� + ����� + �|����||����| = � − ��������� − ����� − ����� ⇔ 
|����||����| + �������� = � 
For �, � ∈ ���� −
�
�
; ��� +
�
�
� or �, � ∈ ���� +
�
�
; ��� +
��
�
� , � ∈ ℤ we get: 
�������� + �������� = � ⇔ ���(� − �) = � ⇔ � − � = ��� ⇔ � = ��� + �, � ∈ ℤ 
 
For � ∈ ���� −
�
�
; ��� +
�
�
� andfor � ∈ ���� +
�
�
; ��� +
��
�
� , � ∈ ℤ we get: 
−�������� + �������� = � ⇔ ���(� + �) = −� ⇔ � + � = (�� + �)� ⇔ 
� = (�� + �)� − �, � ∈ ℤ 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Solution 2 by Ravi Prakash-New Delhi-India 
|����| + |����| = �(� + ���� + ����)(� − ���� − ����); (�) 
⇒ (|����| + |����|)� = � − (���� + ����)� ⇒ 
����� + ����� + �|����||����| = � − ��������� − ����� − ����� ⇔ 
|��������| + �������� = � 
 If �������� ≥ � we get: 
 ���(� − �) = � ⇔ � − � = ��� ⇔ � = ��� + �, � ∈ ℤ ⇒ 
� = ��� + �, � ∈ ℤ 
If �������� ≤ � we get: 
�������� + �������� = � ⇔ ���(� + �) = −� ⇔ � + � = (�� + �)� ⇔ 
� = (�� + �)� − �, � ∈ ℤ 
Thus, � = ��� + �, � ∈ ℤ or � = (�� + �)� − �, � ∈ ℤ 
 Solution 3 by Khaled Abd Imouti-Damascus-Syria 
|����| + |����| = �(� + ���� + ����)(� − ���� − ����) 
|����| + |����| = �� − (���� + ����)� ⇔ 
⇒ (|����| + |����|)� = � − (���� + ����)� ⇒ 
����� + ����� + �|����||����| = � − ��������� − ����� − ����� ⇔ 
� + |��������| = −�������� 
 If �������� > � we get: � + �������� = −�������� ⇔ 
 ���(� − �) = � ⇔ � − � = ��� ⇔ � = ��� + �, � ∈ ℤ 
If �������� < � we get: �������� + �������� = � ⇔ 
���(� + �) = −� ⇔ � + � = (�� + �)� ⇔ � = (�� + �)� − �, � ∈ ℤ 
If �������� = � we get: ���� = � ⇒ � =
�
�
+ �� or ���� = � ⇒ � =
�
�
+ ��, � ∈ ℤ 
150. � = � � ∣∣
∣ � ∈ ℤ, �
����������������
��
� = �, [∗] − ��� �. Find: 
Ω = � �
�∈�
 
Proposed by Daniel Sitaru-Romania 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Solution 1 by Djeeraj Badera-Jaipur-India 
Given that: � ∈ ℤ, �
����������������
��
� = �; (�) 
From def. of GIF: � ≤ �� − ���� + ���� − ��������������������
�(�)
< ��; (�) 
(for satisfying condition A) 
Case 1. 
�� − ���� + ���� − ��� = � ⇔ �(�� − ���� + ���� − ��) = � ⇔ 
�(�� − �)(�� − �)(�� − �) = � ⇔ � ∈ {�, ±�, ±�, ±�} 
Case 2. 
� ∈ ℤ ��� �� − ���� + ���� − ��� < �� 
� ∈ ℤ; � ≥ � 
�(�) = �(�� − �)(�� − �)(�� − �) 
��(�) = (�� − �)(�� − �)(�� − �) + ���(�� − �)(�� − �) + ���(�� − �)(�� − �) +
+ ���(�� − �)(�� − �) 
For all � ∈ ℤ where � ≥ � ⇒ ��(�) > � ⇒ � − increasing. 
�(�) = �(�� − �)(�� − �)(�� − �) > �� ⇒ � ≥ � not satisfy condition (B) similarly for 
� ≤ −�. 
Possible value of � is {�, ±�, ±�, ±�} 
Ω = � �
�∈�
= � + � − � + � − � + � − � = � 
Solution 2 by Ravi Prakash-New Delhi-India 
Let �(�) = �� − ���� + ���� − ��� = �(�� − ���� + ���� − ��) = 
=
����
�(�� − ���� + ��� − ��) = �(�� − �� − ���� + ��� + ��� − ��) = 
�(� − �)(�� − ��� + ��) = �(� − �)(� − �)(� − �) = 
= �(�� − �)(�� − �)(�� − �) = (� + �)(� + �)(� + �)�(� − �)(� − �)(� − �) 
Note that �(�) = � for � ∈ {�, ±�, ±� ± �} ⇒ �
�(�)
��
� = � for � ∈ {�, ±�, ±� ± �} 
⇒ � = {�, ±�, ±� ± �} 
For � ∈ ℤ; � ≥ � 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
�(�) = ��
��� > �
� + �
�
� (�!) ⇒ �(�) > �! = ����, ∀� ≥ � ⇒ �
�(�)
��
� > �, ∀� ≥ � 
Also, for � ≤ −�, � ∈ ℤ �(�) = (−�)���
����� < −�! = −���� ⇒ �
�(�)
��
� < −�, ∀� ≤ −� 
So, Ω = ∑ ��∈� = � 
 Solution 3 by Michael Stergiou-Greece 
� ∈ ℤ, �� =
�� − ���� + ���� − ���
��
� = �; (�) 
First, we observe that: 
� =
�� − ���� + ���� − ���
��
=
(� + �)(� + �)(� + �)�(� − �)(� − �)(� − �)
��
 
If � < −� then � ≤ −�, � ∈ ℤ and ��� < � (odd, that is 7, terms<0) and [�] ≤ −� 
contradiction. 
If � > � then � ≤ −�, � ∈ ℤ and ��� > �! = ���� or � >
����
��
= �� or [�] > � 
contradiction. So, (−� ≤ � ≤ �) ⇒ � ∈ � = {�, ±�, ±� ± �} that satisfy (1) and 
Ω = � �
�∈�
= � 
151. Solve in ℝ the equation: 
����� − ������ − ������� + ������� + ������� − ������� − ������� − ����� + � = � 
Proposed by Rovsen Pirguliyev-Sumgait-Azerbaijan 
Solution by proposer 
����� − ������ − ������� + ������� + ������� − ������� − ������� − ����� + � = � 
⇔ �(����� − �����) = �(� − ������) where �(�) = � + � + �� 
��(�) = � + ��� > � ⇒ � −increasing, then we have: 
�(�) = �(�) ⇔ � = �, then 
����� − ����� = � − ������ ⇔
����� − �����
� − ������
= � ⇔ ����� = � ⇔ 
�� =
�
�
+ �� ⇒ � =
�
��
+
��
�
, � ∈ ℤ 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
152. Solve for �, �, � ∈ �−
�
�
,
�
�
� then: 
���(��)
����
+
���(��)
����
+
���(��)
����
=
��
�
 
Proposed by Daniel Sitaru-Romania 
Solution by Khaled Abd Imouti-Damascus-Syria 
We know that: ���� =
��������
�
; ����� =
�
��
(��� + ����)� 
����� =
�
��
�� �
�
�
�
�
���
���� ∙ ��(���)�� = 
=
�
��
�� ∙
���� + �����
�
+ � ∙ � ∙
���� + �����
�
+ �� ∙ � ∙
��� + ����
�
� = 
=
�
��
[����(��) + �����(��) + ������] 
������� = ����(��) + �����(��) + ������ 
������� = ���(��) + ����(��) + ������ 
���(��) = ������� − ����(��) − ������ 
���(��) = ������� − �(������ − �����) − ������ 
���(��) = ������� − ������� + �����; (���� ≠ �) 
���(��)
����
= ������� − ������� + � = �� ������ −
�
�
�
�
−
�
�
 
So, we get: 
���(��)
����
+
���(��)
����
+
���(��)
����
=
��
�
 
 �� ������ −
�
�
�
�
−
�
�
+ �� ������ −
�
�
�
�
−
�
�
+ �� ������ −
�
�
�
�
−
�
�
=
��
�
 
�� ������� −
�
�
�
�
+ ������ −
�
�
�
�
+ ������ −
�
�
�
�
� −
��
�
=
��
�
 
�� ������� −
�
�
�
�
+ ������ −
�
�
�
�
+ ������ −
�
�
�
�
� =
��
�
 
If ���� = ���� = ���� then: � ∙ �� ������ −
�
�
�
�
=
��
�
⇔ ������ −
�
�
�
�
=
�
��
 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
����� =
�
�
±
√�
�√�
 
Case I. ���� = −�
�
�
∓
√�
�√�
⇒ ���� = −�
�
�
−
√�
�√�
< −� (����������) 
Case II. ���� = �
�
�
∓
√�
�√�
⇒ ���� = �
�
�
−
√�
�√�
∈ (�, �) 
So, � = {�, �, � �⁄ = � = � = ����� ��
�
�
−
√�
�√�
� 
153. Solve for real numbers: 
� ∙ �
�
�
� +
�
�
∙ [�] = �, [∗] − ��� 
Proposed by Marin Chirciu-Romania 
Solution 1 by Bedri Hajrizi-Mitrovica-Kosovo 
� ∙ �
�
�
� +
�
�
∙ [�] = �� : � ⇒ �
�
�
� �� − �� + [�] = � 
∆= � − �[�] ∙ �
�
�
� ≥ � ⇔ [�] ∙ �
�
�
� ≤ �; (∗) 
Case 1. � ∈ (−∞, −�) ∪ (−�, �) ⇒ [�] ∙ �
�
�
� ≥ (−�)(−�) = � contradicts (∗). 
Case 2. � = −�: [−�] ∙ [−�] = � 
[−�] ∙ [−�] + [−�] ∙ [−�] = � ���� ⇒ � = −� −is solution. 
Case 3. � ∈ (�, �): [�] ∙ �
�
�
� = � equation is: � ∙ �
�
�
� = � no solution. 
Case 4. � = � is solution. 
Case 5. � ∈ (�, ∞): [�] ∙
�
�
= � ⇒ �� = [�] no solution. 
Solution 2 by Ravi Prakash-New Delhi-India 
� ∙ �
�
�
� +
�
�
∙ [�] = �; (�) 
First note that equation is defined if � ≠ �. 
Also, if � is solution of (1) then 
�
�
 is also solution of (1), so it is sufficient to consider 
|�| ≥ �. For � = −�, � = �, the equation is satisfied. 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
If � > � ⇒ � <
�
�
< � ⇒ �
�
�
� = � 
Now, (1) can be written as: � ∙ � +
�
�
∙ [�] = � it has no solution as [�] ≤ �. 
Next, let � < −� ⇒ −� <
�
�
< � and [�] = � ≤ −� ⇒ �
�
�
� = −� and [�] = � ≤ −� 
Now, (1) becomes: � = −� +
�
�
= � +
�
�
 where � = −�, � = −� ≥ � ⇒ 
� = � +
�
�
> � +
�
�
≥ � not possible 
So, any solutions are � = ±� 
Solution 3 by Khaled Abd Imouti-Damascus-Syria 
When � > � then 
�
�
< � so �
�
�
� = �. 
So: 
�
�
∙ [�] = � ⇒ [�] = �� ⇒ �� = [�] ≤ � ⇒ � ≤ � impossible. 
When � < −� ⇒
�
�
> −� so �
�
�
� = −�. 
−� − � ≤ � < �; � ∈ ℕ ⇒ [�] = � − � ⇒ −� +
�
�
(� − �) = � ⇔ 
�� + �� + � + � = � with ∆= −�� < �, � ∈ ℕ, no solution. 
When � < � < � ⇒
�
�
> � ⇒ [�] = � equation becomes �
�
�
� =
�
�
 impossible. 
When � = �: � + � = � 
When – � < � < �: � ∙ �
�
�
� −
�
�
= �; ��
�
�
� ≤
�
�
� ⇔ � �
�
�
� ≥ � and � +
�
�
≥ � then � ≤ −� 
impossible. When � = −�: [−�] ∙ [−�] + [−�] ∙ [−�] = � 
� = {−�, �} 
154. Solve for real numbers: 
���� − �� = ����� + � + �� + � + �, � > � − �����. 
Proposed by Marin Chirciu-Romania 
Solution by Santos Martin Junior-Brusels-Belgium 
Let ���� − �� = �; ����� + � = �; �� + � + � = � where �, �, � > � 
Then our equation becomes: � = � + � 
 
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Also: �� − �� + �� = �(��� − �� + �) + (�� + �� + �) ⇔ 
��+ �� − �� = �(�� − �)� + (� + �)�; (∴ � = � + �) ⇔ 
(� + �)� − �� − ��� = �(�� − �)� + (� + �)� ⇔ 
�� − �� − ��� = �(�� − �)� + (� + �)� ⇔ 
−��� = �(�� − �)� + (� + �)� 
��� ≤ � and ��� ≥ � then we must have: ��� = ��� = � 
��� = �: (�, �) = �
�
�
, �� 
��� = �: either � = � ⇔ ���� + � = � ⇔ � − � = � ⇔ � = � or 
� = � ⇔ � + � + � = � ⇔ � = � reject since � > �. 
 
155. Solve for real numbers: 
�� + ��� + ���� + ���� + ���� + ��� = ���, � = ���(����) 
Proposed by H.Tarverdi-Baku-Azerbaijan 
Solution 1 by Bedri Hajrizi-Mitrovica-Kosovo 
Let �(�) = �� + ��� + ���� + ���� + ���� + ��� − ���; (��� = �� ∙ ��) 
By Bezout theorem �(�) = �(−�) = �, we get: 
(� + �)(� − �) �(�)⁄ 
Let �(�) = (� + �)(� − �)�(�) 
�(�)
�(�)
= �� + ��� + ���� + ��� + �� = (�� + ��)� + �(�� + ��) + �� > � 
So, �(�) = (� + �)(� − �)�(�) > � 
Solutions of equation �(�) = �, are 
�� − �; �� = � ⇒ �� = √�
��
; �� = �
� 
Solution 2 by George Florin Şerban-Romania 
�� − �� + ��� − ��� + ���� − ���� + ���� − ���� + ���� − ��� + ���� − ��� = � 
(� − �)��� + ��� + ���� + ���� + ��� + ���� = � 
 � − � = � ⇒ � = � ⇒ ���(����) = � ⇒ �� = �
� 
 
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On the other case, we have: 
 �� + ��� + ���� + ���� + ��� + ��� = � ⇔ 
(� + �)(�� + ��� + ���� + ��� + ��) = � 
 � + � = � ⇒ � = −� ⇒ ���(����) = −� ⇒ �� = �
��� 
 �� + ��� + ���� + ��� + �� = � ⇔ 
 (�� + ��)� + �(�� + ��) + �� > �-no solution in ℝ. 
Solutions of equation are 
�� − �; �� = � ⇒ �� = �
�; �� = �
��� 
 Solution 3 by Orlando Irahola Ortega-La Paz-Bolivia 
�� + ��� + ���� + ���� + ���� + ��� = ��� ⇔ 
��� + ��� + ���� + ���� + ���� + �� + �� + �(�� + ��� + ��� + ��) = ��� ⇔ 
(� + �)� + �(�� + ��� + ��� + �� + �) + �(�� + �� + �) + � = ��� ⇔ 
(� + �)� + �(� + �)� + �(� + �)� + � = ��� ⇔ 
(�� + �� + �)� = �� ⇒ �� + �� + � = � ⇒ (� + �)(� + �) = � ⇒ 
�� − �; �� = � ⇒ �� = �
�; �� = �
��� 
Solution 4 by Abner Chinga Bazo-Lima-Peru 
 
�� + ��� + ���� + ���� + ���� + ��� = ��� ⇔ 
�(� + �)� + �(� + �)� + �(� + �)� − ��� = � ⇔ 
((� + �)� + �)� = �� ⇒ �� + �� + � = � 
 
156. Solve in real numbers: 
� − � = ������ + ���� − ��� − � +
�
�
�
 
Proposed by Orlando Irahola Ortega-La Paz-Bolivia 
Solution by Lety Sauceda-Mexico City-Mexico 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
� − � = ������ + ���� − ��� − � +
�
�
�
⇔ 
(� − �)� = �� ����� + ���� − ��� − � +
�
�
� ⇔ 
���� + ���� − ���� − �� − ��� + �� − � = � 
Let: � = � −
�
�
 then: 
���� − ����� + ����� − ����� + ����� − ���� + �� = � 
���� − ����� + ���� − ��� +
���
�
−
���
��
+
��
��
= � 
�� ��� +
�
��
� − ��� ��� +
�
��
� + ��� �� +
�
�
� − ��� = � 
Let: � = � +
�
�
; �� − � = �� +
�
��
; �� − �� = �� +
�
��
 we get: 
���� − ����� + ���� − ��� = � 
� = � +
���
��
; �� −
���
���
� −
����
�����
= � 
� = � + �; �� + �� −
����
�����
= �; � = ��� −
���
���
= � 
�� −
����
�����
�� −
(���)�
��(���)�
= � 
�� =
����
����� ±
��
����
������
�
− � �
���
� ∙ ����
�
�
 
� + � = � ⇒ � +
���
��
= � 
� =
�
����
����� +
��
����
������
�
− � �
���
� ∙ ����
�
�
�
+
�
����
����� −
��
����
������
�
− � �
���
� ∙ ����
�
�
�
+
���
��
 
� = � +
�
�
; �� − �� + � = � ⇒ � =
� ± ��� − �
�
 
� = �. �������������� … 
� = � −
�
�
= � −
�
�±�����
. So, 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
�� = �. ��������������� … 
�� = −(�. �������������� 
Solution 2 by Arslan Ahmed-Sanaa-Yemen 
� − � = ������ + ���� − ��� − � +
�
�
�
⇔ 
�
�
�
− ��
�
= ����� + ���� − ��� − � +
�
�
� ⇔ 
−�� + ��� − �� + �
��
= ���� + ���� − ��� − � −
�
�
⇔ 
−�� + ��� − �� + � = ���� + ���� − ���� − ��� − �� ⇔ 
�� +
��
��
�� −
�
�
�� +
�
��
�� −
�
�
�� +
�
�
� + � =
��
��
 
Now, �� +
��
��
�� −
�
�
�� +
�
��
�� −
�
�
�� +
�
�
� + � = (� + �)� 
Sum of the boundaries � + � = � ⇒ � = � 
�� = �
�
�
� ���� = �� ⇒ � = � 
�� = �
�
�
� ���� =
��
��
�� ⇒ � =
��
���
 
�� +
��
���
�
�
=
��
��
 
�� = �
��
��
�
−
��
���
; �� = −�
��
��
�
+
��
���
 
�� = −�. �����; �� = �. ������ 
 
157. Solve for real numbers: 
�� + �
�� + � + �
+
�� + �
�� + � + �
+
�� + �
�� + � + �
+ � = � 
Proposed by Daniel Sitaru-Romania 
Solution 1 by George Florin Şerban-Romania 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
�
�� + �
�� + � + �
���
+ � = � ⇒ � �
�� + �
�� + � + �
+
�
�
�
���
= � ⇒ 
�
��� + �� + �
�� + � + �
���
= � ⇒ �
(�� + �)�
�� + � + �
���
≥ �, ∀� ∈ ℝ, 
(�� + �)� ≥ � and ⇒ �� + � + � > 0, ∆= −3 < 0 
�
(�� + �)� = �
(�� + �)� = �
(�� + �)� = �
⇒ � = � = � = −
�
�
 
Solution 2 by Florică Anastase-Romania 
Let’s denote �
� = �� + � + �
� = �� + � + �
� = �� + � + �
 
� = �� + � + � = �� +
�
�
�
�
+
�
�
≥
�
�
 and similarly � ≥
�
�
, � ≥
�
�
 
Hence, 
�
�� + �
�� + � + �
���
+ � = � ⇔ �
� − �
�
���
+ � = � 
� �
� − �
�
+
�
�
�
���
= � ⇔ �
�� − �
��
���
= � ⇔ 
� = � = � =
�
�
⇔ � = � = � = −
�
�
 
Solution 3 by Nikos Ntorvas-Greece 
Let be the function �(�) =
����
������
, � ∈ ℝ, �� + � + � > � 
��(�) =
�� + �
(�� + � + �)�
 
� 
−∞ −
�
�
 + ∞ 
��(�) − − − − − − −� + + + + + + + + 
�(�) 
↘ ↘ ↘ ↘ � �−
�
�
� ↗ ↗ ↗ ↗ 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
Hence, �(�) ≥ � �−
�
�
� = −
�
�
, ∀� ∈ ℝ. So, we have that: �(�) ≥ −
�
�
, ∀� ∈ ℝ. 
Equality holds for � = −
�
�
. For �, �, � ∈ ℝ we have: 
�(�) + �(�) + �(�) ≥ � �−
�
�
� ⇔ �(�) + �(�) + �(�) ≥ −� ⇔ 
�(�) + �(�) + �(�) + � ≥ �. 
Equality holds if and only if � = � = � = −
�
�
. 
158. Solve for real numbers: 
�[����] = [�], [∗] −is the greatest integer parts of ∗ 
Proposed by Jalil Hajimir-Toronto-Canada 
Solution by Bedri Hajrizi-Mitrovica-Kosovo 
�[����] = [�]; (�) 
i) [����] = −�
(�)
�� −� = [�] ⇒ � + [�] = � 
For � < � ⇒ � + [�] < � and for � ≥ � ⇒ � + [�] = � ⇒ � = � which is impossible. 
ii) [����] = �
(�)
�� [�] = � ⇒ � ∈ [�, �) and ���� ∈ [�, �) ⇒ � ∈ ��,
�
�
� ⇒ � ∈ [�, �) 
iii) [����] = �
(�)
�� � = [�] ⇒ � ∈ ℤ and ���� = � ⇒ � =
�
�
+ ���, � ∈ ℤ ⇒ � ∈ ∅ 
Finally, � ∈ [�, �) 
159. If � ∈ ��,
�
��
� , � − fixed, solve for real numbers: 
�����(� − �) + ����(�� − � + �) + �(�� − �) + �(� − �) = � 
Proposed by Marin Chirciu-Romania 
Solution by George Florin Şerban-Romania 
�����(� − �) + ����(�� − � + �) + �(�� − �) + �(� − �) = � ⇔ 
����� − ����� + ���� − ���� + ���� + (�� − ��)� + � − �� = � 
(�� − �)������ − ����� + (� − ��)�� − (�� + �)�� + (� − ��)� − �� − �� = � 
i) �� − � = � ⇒ �� =
�
�
 
ii) ����� − ����� + (� − ��)�� − (�� + �)�� + (� − ��)� − �� − � = � ⇔ 
 
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 �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 
 
(�� − �)[����� + (� − ��)�� − ��� + � − ��] = � ⇒ �� =
�
�
 
iii) ����� + (� − ��)�� − ��� + � − �� = � 
If � = � ⇒ ��� + � = �, ∆= −�� no solution in ℝ. 
If � ≠ �, ∆= ��[�� − �(� − �)�] < � because � ∈ ��,
�
��
� 
We have that: 
�� − �(� − �)� < � ⇒ �� < �(� − �)� ⇒ |�| < √�|� − �| ⇒ � < √�(� − �) 
��� + √�� < √� ⇒ � <
√�
��√�
 true because � ≤ � ≤
�
��
<
√�
��√�
. 
Hence, ����� + (� − ��)�� − ��� + � − �� > �. So, � = �
�
�
�. 
 
160. Solve for real numbers: 
��√� + �
�
− ��√� − �
�
= √�
�
 
Proposed by Daniel Sitaru-Romania 
Solution by Santos Martins Junior-Brusels-Belgium 
Let � = �√� + �; � = �√� − � 
Then � − � = √�
�
; �� − �� = ��; �� + �� = ��√� 
Then 
�
√�
� −
�
√�
� = �; �
�
√�
� �
�
− �
�
√�
� �
�
= �; �
�
√�
� �
�
+ �
�
√�
� �
�
= �√� 
Let 
�
√�
� = �; 
�
√�
� = � 
We have: � − � = �; (�), �� − �� = �; (�), �� + �� = �√�; (�) 
Also, we know: �� + �� = (� + �)(�� − ��� + ���� − ��� + ��) 
= (� + �)[(�� − ��)� + ����� − ��(�� − �� + ��)] 
= (� + �)[(� − �)�(� + �)� + ����� − ��((� − �)� + ��)] 
= (� + �)[(� − �)�((� − �)� + ���) + ����� − ��((� − �)� + ��)] 
Now, �√� = (� + �)[� ∙ (� + ���) + ����� − ��(� +

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