Inequality with $abc=1$,

Question

Given 3 positive numbers $a, b, c$ satisfying $abc= 1$

Prove: $$\frac{1}{a^{5}+ b^{5}+ c^{2}}+ \frac{1}{b^{5}+ c^{5}+ a^{2}}+ \frac{1}{c^{5}+ a^{5}+ b^{2}}\leq 1$$

My opinion:

Let: $x= \frac{a}{b}, y= \frac{b}{c}, z= \frac{c}{a}$

We have to prove:

$$\sum \frac{x^{2}y^{5}z^{5}}{x^{7}z^{5}+x^{2}y^{10}+y^{5}z^{7}}\leq 1$$

I want to prove:

$$\frac{x^2y^5z^5}{x^7z^5+x^2y^{10} +y^5z^7} \leq \frac{\frac{x^2}{z^2}}{2\sqrt{\frac{x^9}{z^9}}+1}$$

Help me, please! Thanks!

Answer 1

By Muirhead we obtain: $$\sum_{cyc}\frac{1}{a^5+b^5+c^2}\leq\sum_{cyc}\frac{1}{a^4b+ab^4+c^3ab}=\sum_{cyc}\frac{1}{ab(a^3+b^3+c^3)}=\frac{a+b+c}{a^3+b^3+c^3}\leq1.$$ For the proof of the last inequality we can use also AM-GM: $$a^3+b^3+c^3=\frac{1}{9}\sum_{cyc}(5a^3+2b^3+2c^3)\geq\frac{1}{9}\sum_{cyc}9\sqrt[9]{\left(a^3\right)^5\left(b^3\right)^2\left(c^3\right)^2}=$$ $$=\sum_{cyc}\sqrt[3]{a^5b^2c^2}=\sum_{cyc}\sqrt[3]{a^3}=a+b+c.$$