I was trying to prove the inequality : for a,b,c positive real numbers where $abc=1$ prove $$\frac{1}{a^{5}+b^{5}+c^{2}}+\frac{1}{b^{5}+c^{5}+a^{2}}+\frac{1}{c^{5}+a^{5}+b^{2}}\leq 1 . $$ It is easy to prove it with standard methods like AM-GM and Cauchy , but I tried it with calculus. Here is my solution: W.L.O.G let $a=\max{a,b,c}$ . Consider the function $f(c)=\frac{1}{a^{5}+b^{5}+c^{2}}+\frac{1}{b^{5}+c^{5}+a^{2}}+\frac{1}{c^{5}+a^{5}+b^{2}}-1$ , clearly $$f'(c)=-\frac{2c}{(a^{5}+b^{5}+c^{2})^2}-\frac{5c^4}{(b^{5}+c^{5}+a^{2})^2}-\frac{5c^4}{(c^{5}+a^{5}+b^{2})^2}$$ wich is clearly negative.Therefore $f$ is decreasing , $f(c)\leq f(a)=g(b)$, and again in the same way we can show that $g(b)$ decreases, now we have $$g(b)\leq g(a) \Longleftrightarrow 2a^5+a^2\geq 3$$ wich is obvious since $a\geq 1$ .

But , I really have doubts this solution works , like can I just consider the inequality like a one variable function, and then like an another function of another variable. Thanks to anyone who responds.

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    $\begingroup$ The problem is that if you let $c$ decrease and keep $a, b$ fixed, the condition $abc=1$ is violated, so I would say no, this is not valid reasoning. $\endgroup$ – saulspatz May 4 at 16:21
  • $\begingroup$ Thanks, this was my reasoning, too , but i wasn't sure. $\endgroup$ – Arben_Ajredini May 4 at 17:29
  • $\begingroup$ another question, so if there would be no condition on $a,b,c$ would the solution hold? $\endgroup$ – Arben_Ajredini May 4 at 19:03
  • $\begingroup$ What if $a=b=c=.00001$? $\endgroup$ – saulspatz May 4 at 19:24
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    $\begingroup$ It's legitimate to hold some variables fixed and let others vary. I'm not sure I follow exactly what the rest of your reasoning is, so I can't say if it's correct. $\endgroup$ – saulspatz May 4 at 19:40

By Muirhead twice we obtain: $$\sum_{cyc}\frac{1}{a^5+b^5+c^2}\leq\sum_{cyc}\frac{1}{a^4b+ab^4+c^3ab}=\sum_{cyc}\frac{c}{a^3+b^3+c^3}=\frac{a+b+c}{a^3+b^3+c^3}\leq1.$$


Another way:

By C-S and Muirhead we obtain: $$\sum_{cyc}\frac{1}{a^5+b^5+c^2}=\sum_{cyc}\frac{\frac{1}{a}+\frac{1}{b}+c^2}{(a^5+b^5+c^2)\left(\frac{1}{a}+\frac{1}{b}+c^2\right)}\leq\sum_{cyc}\frac{\frac{1}{a}+\frac{1}{b}+c^2}{(a^2+b^2+c^2)^2}=$$ $$=\sum_{cyc}\frac{bc+ac+c^2}{(a^2+b^2+c^2)^2}=\left(\frac{a+b+c}{a^2+b^2+c^2}\right)^2\leq1.$$


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