florin_rotaru wrote: Problem Shortlist BMO 2017 ( Proposed by Florin Rotaru Romania) Let $ a $,$ b$,$ c$, be positive real numbers such that $abc= 1 $. Prove that $$\frac{1}{a^{5}+b^{5}+c^{2}}+\frac{1}{b^{5}+c^{5}+a^{2}}+\frac{1}{c^{5}+b^{5}+b^{2}}\leq 1 . $$ By Muirhead $$\sum_{cyc}\frac{1}{a^5+b^5+c^2}\leq\sum_{cyc}\frac{1}{a^4b+b^4a+c^2}=\sum_{cyc}\frac{c}{a^3+b^3+c^3}\leq1.$$

Note that, $a^5+b^5\geq a^4b+ab^4$. With this, the left hand side is upper bounded by $\sum_{cyc}\frac{1}{a^4b+ab^4+c^3ab}$, using $abc=1$. Hence, the inequality boils down to $$ \sum_{cyc}\frac{1}{a^4b+ab^4+c^3ab} = \sum_{cyc}\frac{c}{a^3+b^3+c^3}\leq 1, $$namely, it boils down showing $a^3+b^3+c^3\geq a+b+c$. Now, using power mean inequality, we have $a^3+b^3+c^3\geq \frac{(a+b+c)^3}{9}$, which, together with $a+b+c\geq 3\sqrt[3]{abc}=3$, yields the proof.

florin_rotaru wrote: Problem Shortlist BMO 2017 ( Proposed by Florin Rotaru Romania) Let $ a $,$ b$,$ c$, be positive real numbers such that $abc= 1 $. Prove that $$\frac{1}{a^{5}+b^{5}+c^{2}}+\frac{1}{b^{5}+c^{5}+a^{2}}+\frac{1}{c^{5}+b^{5}+b^{2}}\leq 1 . $$

florin_rotaru wrote: Problem Shortlist BMO 2017 ( Proposed by Florin Rotaru Romania) Let $ a $,$ b$,$ c$, be positive real numbers such that $abc= 1 $. Prove that $$\frac{1}{a^{5}+b^{5}+c^{2}}+\frac{1}{b^{5}+c^{5}+a^{2}}+\frac{1}{c^{5}+b^{5}+b^{2}}\leq 1 . $$ https://artofproblemsolving.com/community/c6h601042p10345751

My solution of April 2017 We have $$ a^{5}+b^{5}\geq ab(a^{3}+b^{3})\Leftrightarrow a^{5}-a^{4}b-ab^{4}+b^{5}\geq 0\Leftrightarrow a^{4}(a-b)-b^{4}(a-b)\geq 0\Leftrightarrow $$$$(a-b)(a^{4}-b^{4})\geq 0\Leftrightarrow (a-b)^{2}(a^{2}+b^{2})(a+b)\geq 0 $$, that is true. Inequality can write: $$ \frac{1}{a^{5}+b^{5}+abc^{3}}+\frac{1}{b^{5}+c^{5}+bca^{3}}+\frac{1}{c^{5}+b^{5}+cab^{3}}\leq 1 . $$Also $$a^{5}+b^{5}+abc^{3}\geq ab(a^{3}+b^{3}+c^{3})$$and analogs. Then $$ \frac{1}{a^{5}+b^{5}+abc^{3}}+\frac{1}{b^{5}+c^{5}+bca^{3}}+\frac{1}{c^{5}+b^{5}+cab^{3}}\leq \frac{1}{a^{3}+b^{3}+c^{3}}(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})=$$$$\frac{a+b+c}{a^{3}+b^{3}+c^{3}}\leq \frac{a+b+c}{\frac{(a+b+c)^{3}}{9}}=\frac{9}{(a+b+c)^{2}}\leq \frac{9}{(3\sqrt[3]{abc})^{2}}=1 . $$

Haven't done an inequality in so long... By Cauchy-Schwarz Inequality, \[(x^5 + y^5 + z^2)\left(\frac{1}{x} + \frac{1}{y} + z^2\right) \ge (x^2 + y^2 + z^2)^2 \iff \frac{1}{x^5 + y^5 + z^2} \le \frac{z(x + y + z)}{(x^2 + y^2 + z^2)^2}\]Summing up over $a, b, c$, we get \[\sum_{cyc} \frac{1}{a^5 + b^5 + c^2} \le \frac{(a + b + c)^2}{(a^2 + b^2 + c^2)^2}.\]Thus, it suffices to prove that \[a^2 + b^2 + c^2 \ge a + b + c = \sqrt[3]{abc}(a + b + c) \iff \sum_{cyc} \frac{4a^2 + b^2 + c^2}{6} \ge \sum_{cyc} \sqrt[3]{a^4bc}\]which is true by AM-GM.

Vrangr wrote: Haven't done an inequality in so long... By Cauchy-Schwarz Inequality, \[(x^5 + y^5 + z^2)\left(\frac{1}{x} + \frac{1}{y} + z^2\right) \ge (x^2 + y^2 + z^2)^2 \iff \frac{1}{x^5 + y^5 + z^2} \le \frac{z(x + y + z)}{(x^2 + y^2 + z^2)^2}\]Summing up over $a, b, c$, we get \[\sum_{cyc} \frac{1}{a^5 + b^5 + c^2} \le \frac{(a + b + c)^2}{(a^2 + b^2 + c^2)^2}.\]Thus, it suffices to prove that \[a^2 + b^2 + c^2 \ge a + b + c = \sqrt[3]{abc}(a + b + c) \iff \sum_{cyc} \frac{4a^2 + b^2 + c^2}{6} \ge \sum_{cyc} \sqrt[3]{a^4bc}\]which is true by AM-GM. Yes,it's exactly post #4 . https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYy9lLzI4MjFmM2Q0Nzg3Y2MzMDkxYjBkOWQ0MDNjYmY2YzA3N2I4MjE5LnBuZw==&rn=eHh5LlBORw==

Solutia:

the second page:

A problem by professor Marian Cucoanes: If $abc=1$ for the real positive numbers $a, b, c$, prove that: $ \frac{1}{a^{\sqrt{2}}+b^{\sqrt{2}}+c}+\frac{1}{b^{\sqrt{2}}+c^{\sqrt{2}}+a} + \frac{1}{c^{\sqrt{2}}+a^{\sqrt{2}}+b} \le 1 \ \ ; $ Greetings!

Hea! See here the generalization from the professor.The Rotaru's problem was subtle. Buut this one... Just a post.

Thank you,professor Teomihai for you multiple appreciations and your kindness!

with pleasure,you have only very nice solutions.

A problem by professor Marian Cucoanes: Let $a, b, c >0$ with $abc=1$ and $ m>n>0$ real numbers. Prove that: $ \frac{1}{a^m+b^m+c^n}+\frac{1}{b^m+c^m+a^n}+\frac{1}{c^m+a^m+b^n} \le 1 \ \ ; $ Greetings!

Hea! By a brilliant idea of mister Arqady,we are done. So you have to work hard.Try to refine my number theory problem,or even BMO 2019 problem.

A problem by professor Marian Cucoanes: Let $a, b, c >0$ with $abc=1$, real numbers. Prove that: $ \frac{1}{a^5+b^2+c^2}+ \frac{1}{b^5+c^2+a^2} +\frac{1}{c^5+a^2+b^2} \le 1 \ \ ; $ Greetings!

Hea! @Above is a pathetic attempt of a so called author,which signs on my idea from posts #4 and #13,with Cauchy. Out of respect for the topic and Rotaru,who indeed is an ORIGINAL creator,I'll comment no more on this or on spam's quality,but I'll kill with an intelligent solution.

Thank you all and and I apologize for not responding to everyone . It may be that Professor Song Qing has found the inequality of my own, I have no other explanation, it is certain that I sent this proposal in April 2017 and I received the answer that she was accepted in the shortlist for the Balkan Olympics seniors 2017. My only testimony is the leader of Balcaniadei senior mathematics team since 2017.

mihaig wrote: florin_rotaru wrote: Problem Shortlist BMO 2017 ( Proposed by Florin Rotaru Romania) Let $ a $,$ b$,$ c$, be positive real numbers such that $abc= 1 $. Prove that $$\frac{1}{a^{5}+b^{5}+c^{2}}+\frac{1}{b^{5}+c^{5}+a^{2}}+\frac{1}{c^{5}+b^{5}+b^{2}}\leq 1 . $$ This is probably the most straight forward solution. Nice!