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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 www.ssmrmh.ro � 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 www.ssmrmh.ro � 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 www.ssmrmh.ro � 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 www.ssmrmh.ro � 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 www.ssmrmh.ro � 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, � = �. www.ssmrmh.ro � 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: www.ssmrmh.ro � 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. www.ssmrmh.ro � RMM-ABSTRACT ALGEBRA MARATHON 101-200 107. If � ∈ ��(ℝ), ���� = �, ���(� � + � + ��) = � then find: Ω = ��� 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: Ω = ���� Proposed by Marian Ursărescu-Romania www.ssmrmh.ro � 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: �� + �√� − �√� = �. www.ssmrmh.ro �� 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 www.ssmrmh.ro �� 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 ��� � ���� ���� � ≥ �� + � � � � ��� � ���� ����√��� � ⇔ ��� � ���� ���� � − ��� � ���� ����√��� � ≥ � � � ��� � ���� ����√��� � ⇔ ��� � ���√�� ���� � ≥ � � � ��� � ���� ����√��� � www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 ⇔ √��������√��� − √����(����) ≥ √����(����) − √��������√��� ⇔ �√� + √���������√��� ≥ √����(����) + √����(����) ⇔ �������√��� ≥ √� √� + √� ���(����) + √� √� + √� ���(����); (�) We have: √� √��√� ���(����) + √� √��√� ���(����) ≤ ������������ ≤ ��� � √� √� + √� ���� + √� √� + √� ����� ≤ ������������ (�,�) ≤ ��� ���� � �√� + �√� √� + √� �� = ��� �����√���� ; (�) From (1),(2) it follows that: ��� � ���� ���� � ≥ �� + � � � � ��� � ���� ����√��� � 113. If � < � < � < � and �, �, � > � Ω� = �� + �� � � � �� + �� � � − � � + �� + �� � � � �� + �� � � − � � + �� + �� � � � �� + �� � � − � � 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. www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 Ω� > (� + �)(� + �) + (� + �)(� + �) + (� + �)(� + �) = [� + (� + �) + ��] + [� + (� + �) + ��] + [� + (� + �) + ��] ≥ ��������� ≥ ��� + (� + �) + �� + �� + (� + �) + �� + �� + (� + �) + ��� � � + � + � ≥ But �� + (� + �) + �� ≥ ��������� � + ��� � + �� � and analogs. ≥ �� + �� + �� + �� + �� + �� + �� � � � � = = � �� + �(� + � + �) + �� + �� + �� � � � � ≥ ��� = � �� + ��√���� � � � = �Ω� Solution 2 by George Florin Şerban-Romania �� + �� � � � �� + �� � � − � � ≥ ������� �� + ��� � � ∙ �� � � − � � � = = �� + � ��� ��(� − �) � � ≥ ��� �� + � ��� � + � − � � � � = �� + √���� � Similarly: �� + �� � � � �� + �� � � − � � ≥ �� + √���� � �� + �� � � � �� + �� � � − � � ≥ �� + √���� � ��� + √���� � ��� ≥ � �� + ��√���� � � � ⇔ www.ssmrmh.ro �� 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, � �� + �� � � � �� + �� � � − � � > ��� ��� + √���� � ��� ≥ ��� ≥ ��� ���� + √���� � �� + √���� � �� + √���� �� = = � ��� + √���� � ��� + √���� � ��� + √���� � �� � ≥ ������� www.ssmrmh.ro �� 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: Ω = �� + � Proposed by Rajeev Rastogi-India Solution 1 by Santos Martins Junior-Brusels-Belgium Let �� + �� − �� + � = ��; (�) �� + �� + � = ��; (�) �� + �� − �� − � = ��; (�)�� + �� + �� − � = ��; (�), where �, �, �, � ≥ � Hence equation becomes: � + � = � + �; (�) www.ssmrmh.ro �� 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, Ω = �� + � = � Solution 2 by Abner Chinga Bazo-Peru ��� + �� − �� + � + ��� + �� + � = ��� + �� − �� − � + ��� + �� + �� − � ��� + �� − �� + � − ��� + �� − �� − � = ��� + �� + �� − � − ��� + �� + � �������������������������������������������������� ����������������������� = = �������������������������������������������� �������������������� −(�� + � − �) ��� + �� − �� + � + ��� + �� − �� − � = �� + � − � ��� + �� + �� − � + ��� + �� + � �� + � − �. Hence, Ω = �� + � = � 116. If � ∈ ℂ, � ∈ [�, ��), |�| = � then: ��� + � − ���� + ������ + ����� + ����� − �� + �� + ����� + ����� + �� − �� ≤ � Proposed by Daniel Sitaru-Romania www.ssmrmh.ro �� 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� |�� − ��| ≤ � |�� − ��| ≤ � |�� − ��| ≤ � ⇒ |�� − ��| + |�� − ��| + |�� − ��| ≤ �; (�) www.ssmrmh.ro �� 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: �� � + � + �� � + � + �� � + � = � �� � + � + �� � + � + �� � + � ≤ � + � � + � + � � + � + � � = � + � + � � www.ssmrmh.ro �� 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: (�(� + �) + ��)� = ���(� + �)��(� + �)� ((� + �)� + ��)� = ���(� + �)� (� + �)� + ���� + ���(� + �)� = ���(� + �)� (� + �)� − ���(� + �)� + ���� = � ((� + �)� − ��)� = � (� + �)� = � ⇔ �(� − �) + (� + �)� � = (� − �)(� + �) ⇔ (�� + �)� = (�� + � − �) ⇔ �(�� + � + �) = � ⇔ ⎩ ⎪ ⎨ ⎪ ⎧ �� = − � � + �√� � �� = − � � − �√� � ⇒ |��| = |��| = �. www.ssmrmh.ro �� 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 www.ssmrmh.ro �� 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 � ��� + �� ��� + � � � ��√�� � + � ��� ≥ ��� √� � √�� ��� + � √� � √�� ��� = √� ∙ � + �√� ∙ � = �√� www.ssmrmh.ro �� 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: (� + �)(� + �)(� + �) ≥ ���� �(� + �)(� + �)(� + �) ≥ ��(� + �)(� + �)(� + �) + ���� = = �(� + � + �)(�� + �� + ��) � �(�� + �� + ��) ≥ �(� + � + �) (� + �)(� + �)(� + �) = (� + �) + (� + �) + (� + �) (� + �)(� + �)(� + �) = = � � (� + �)(� + �) ��� ���� �(�� + �� + ��) ≥ (� + � + �)��� � � (� + �)(� + �) ��� ≥ ������ �� � �(� + �)(� + �) � ��� � � www.ssmrmh.ro �� 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, �(��� + ���) ≥ (�� + ��)(�� + ��) �(��� + ���) ≥ (�� + ��)(�� + ��) �(��� + ���) ≥ (�� + ��)(�� + ��) www.ssmrmh.ro �� 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: � (����) ������ � � ≤ � �� ; (�������) ↔ (�� + �) ���� + � ≤ � � ; ↔ ��� + �� ≤ ���� + �; ↔ ���� − ��� − �� + � ≥ �; ↔ (�� + �)(�� − �)� ≥ �; www.ssmrmh.ro �� 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: www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 �� ≔ � � − � ���������������(���) > � 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: ����� + ����� + ����� � ≥ � � + � + � � ��� � + � + � � � = � � ��� � � www.ssmrmh.ro �� 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 (��� + ��� + ���)(��� + ��� + ���)(��� + ��� + ���) = www.ssmrmh.ro �� 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. www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 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 � √�� ��� + √�� ��� + √�� ��� � ��� ≤ ��������� ����(�� + �� + ��) www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 �� + �� + �� ≥ � √�� ��� + √�� ��� + √�� ��� � ��� ���� �� + �� + �� ≥ (�) � √��� ��� � √� ��� + √� ��� + √� ��� �� ��� ���� (�� + �� + ��)��� ≥ (���)��� �� √� ��� + √� ��� + √� ��� � ��� � ��� ����∙��� ≥ ��� ≥ ��� (���) ��� ���� √��� ���� � ��� � √� ��� + √� ��� + √� ��� �� ��� ����∙��� = (���)��� �����(���) �� ���� √� ��� + √� ��� + √� ��� �� ��� ����∙��� = ����∙���(���)���(���)��� √� ��� + √� ��� + √� ��� � ��� ����∙��� = (���)���� √� ��� + √� ��� + √� ��� � ��� Therefore, (��������)��� (���)��� ≥ � √� ��� + √� ��� + √� ��� � ��� Solution 2 by Sanong Huayrerai-Nakon Pathom-Thailand For �, �, � > 0 we give: � = ����, � = ����, � = ���� Hence, (��������)��� (���)��� ≥ � √� ��� + √� ��� + √� ��� � ��� �(��)��� + (��)��� + (��)���� ��� ((���)���)��� ≥ (� + � + �)��� (��)��� + (��)��� + (��)��� (���)��� ≥ � + � + � (��)��� ���� + (��)��� ���� + (��)��� ���� ≥ � + � + � ((��)��� + (��)��� + (��)���)� (��)������� + (��)������� + (��)������� ≥ � + � + � (��)��� + (��)��� + (��)��� + �(������������ + ������������ + ������������) ≥ ≥ (� + � + �)((��)������� + (��)������� + (��)�������) www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 (��)��� + (��)��� + (��)��� + ������������ + ������������ + ������������ ≥ ≥ ������������ + ������������ + ������������ + ������������ + ������������ + ������������ 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 www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 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 (� + � + �)� �� + �� + �� = (� + � + �)�(� + � + �)� �� + �� + �� = = �� + �� + �� + ��� + ��� = ��� �� + �� + �� ∙ (� + � + �)� = www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 = � �� + �� + �� �� + �� + �� + �� (� + � + �)� ≥ (� + �) ∙ ����� = �� 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 (�����)� �������� ≥ �� ⇔ ����� �������� ∙ � ����� � � � ≥ � www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 ⎝ ⎜⎜ ⎛ � + � � � + � + � �� + �� + �� + � � + � + � � ⎠ ⎟⎟ ⎞ � ≥ � � �� + �� + �� � + � + � + �� � + � + � ≥ � ⇔ �(� + � + �) ≥ �� + �� + �� + �� 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: www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 � � � � ��(�) � ����∈�� ≥ �! ∙ �� � �� ��(�) � ����∈�� �! = (�� ������ ��, � = �, ������, ��� ������� ������� ��� �� �������� ��(� − �)! �����) = �! ∙ ��� (���)! ∙ �� (���)! ∙ … ∙ �� (���)!�! = �! ∙ ��� �� ���� � (���)! �! = �! ∙ �� �� � ��� � 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 �(�) = � � ��� + � � � ����, � > � ��(�) = � �� (� − ����) + � � + � � � � − ���� � + � � ; (�) www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 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 www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 �(� + � + �)� ≥ ��(��� + ��� + ��� + ���); (�) � ≥ � ≥ � ⇒ (� − �)(� − �) ≥ � ⇒ �� + �� ≤ �� + �� (�) �� ��� + ��� + ��� + ��� = ��� + �(�� + ��) + ��� ≤ ≤ ��� + �(�� + ��) + ��� = �(� + �)� = = � � ∙ ��(� + �)(� + �) ≤ � � ∙ � �� (�� + � + � + � + �)� 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 (� − �)� ��� + ��� � � � � ≤ � − � � �� � � � − �� � ��� + ��� � � ≤ � � − � www.ssmrmh.ro �� 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: www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 ��� � � � � ≤ � � �� � − � �� � ⇔ ��� � � � � ≤ �� − �� ��� (� − �)� ��� + ��� � � � � ≤ (� − �)� ��� + �� − �� ��� 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, �� + � = � ����� www.ssmrmh.ro �� 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 � = � � ∈ (�, �] ⇔ � = ��. �� �� ≥ ��√��� ��� ⇔ ��� �� �� ≥ �����√��� ��� ⇔ ����� − ����� ≥ (� − �) �� + � � ���(��)� ����� − (��)���(��) ≥ (� − ��) �� + ���(��) + ���� � � www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 ����� − ������ − ������ ≥ �(� − �) �� + ���� + ���� � � ���� ≤ �� − � � + � , ∀� ∈ (�, �]; (�) 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 � ≥ � ⇒ �(�) ≥ �(�) = �; ∀� ≥ � ���� ≥ ��(���); ∀� ≥ � www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 � � � � �� � � ≥ ��( � � ��); � � ≥ � ⇒ (�) 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�|(� − �)� + (� − ��)� ≤ ��} ⊊ www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 {(�; �) ∈ 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 www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 (� − �)‼ (� − �)‼ (� − �)‼ (� − �)‼ + (� − �)‼ (� − �)‼ (� − �)‼ (� − �)‼ + (� − �)‼ (� − �)‼ (� − �)‼ (� − �)‼ = �� ⇔ (� − �)(� − �) + (� − �)(� − �) + (� − �)(� − �) = �� ⇔ ��� − ��� = � ⇔ ��(� − �) = �, � ∈ 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, (�� − �) ∙ � (����) = ��� − �� (�� + �)(��� − ��) = �� − � ���� − ��� − ��� = � www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 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 � = � ⇒ � = √� � � + � = √� � ≠ � www.ssmrmh.ro �� 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, � + � = √� � ⇒ (� + �)� = � = �� + �� ⇒ � � � � ��� + � � � � ���� + � � � � ���� + � � � � ���� + � � � � ���� + � � � � ��� = � ������� + ��� + ���(�� + ��) + �����(� + �)� = � www.ssmrmh.ro �� 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, ��, ��, … , �� ∈ [−�, �] www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 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 www.ssmrmh.ro �� 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 www.ssmrmh.ro �� 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; � ≥ � www.ssmrmh.ro �� 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 www.ssmrmh.ro �� 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: � ∙ �� ������ − � � � � = �� � ⇔ ������ − � � � � = � �� www.ssmrmh.ro �� 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. www.ssmrmh.ro �� 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: � = � + � www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 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 �� − �� + ��� − ��� + ���� − ���� + ���� − ���� + ���� − ��� + ���� − ��� = � (� − �)��� + ��� + ���� + ���� + ��� + ���� = � � − � = � ⇒ � = � ⇒ ���(����) = � ⇒ �� = � � www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 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 www.ssmrmh.ro �� RMM-ABSTRACT ALGEBRA MARATHON 101-200 � − � = ������ + ���� − ��� − � + � � � ⇔ (� − �)� = �� ����� + ���� − ��� − � + � � � ⇔ ���� + ���� − ���� − �� − ��� + �� − � = � Let: � = � − � � then: ���� − ����� + ����� − ����� + ����� − ���� + �� = � ���� − ����� + ���� − ��� + ��� � − ��� �� + �� �� = � �� ��� + � �� � − ��� ��� + � �� � + ��� �� + � � � − ��� = � Let: � = � + � � ; �� − � = �� + � �� ; �� − �� = �� + � �� we get: ���� − ����� + ���� − ��� = � � = � + ��� �� ; �� − ��� ��� � − ���� ����� = � � = � + �; �� + �� − ���� ����� = �; � = ��� − ��� ��� = � �� − ���� ����� �� − (���)� ��(���)� = � �� = ���� ����� ± �� ���� ������ � − � � ��� � ∙ ���� � � � + � = � ⇒ � + ��� �� = � � = � ���� ����� + �� ���� ������ � − � � ��� � ∙ ���� � � � + � ���� ����� − �� ���� ������ � − � � ��� � ∙ ���� � � � + ��� �� � = � + � � ; �� − �� + � = � ⇒ � = � ± ��� − � � � = �. �������������� … � = � − � � = � − � �±����� . So, www.ssmrmh.ro �� 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 www.ssmrmh.ro �� 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, �� + � + � > � ��(�) = �� + � (�� + � + �)� � −∞ − � � + ∞ ��(�) − − − − − − −� + + + + + + + + �(�) ↘ ↘ ↘ ↘ � �− � � � ↗ ↗ ↗ ↗ www.ssmrmh.ro �� 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) ����� − ����� + (� − ��)�� − (�� + �)�� + (� − ��)� − �� − � = � ⇔ www.ssmrmh.ro �� 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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