($x,y,z$ are positive real numbers)

Question

I need this for lemma but I don't know how to prove it. Maybe rearrangement inequality or arithmetic-geometric mean inequality would work.

Answer 1

By Cauchy-Schwarz's inequality:

$$\left(x^5+y^2+z^2\right)\left(\frac{1}{x}+y^2+z^2\right)\ge \left(x^2+y^2+z^2\right)^2$$

$$\Rightarrow \text{L.H.S}=\sum _{cyc}\frac{1}{x^5+y^2+z^2}\le \frac{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+2\left(x^2+y^2+z^2\right)}{\left(x^2+y^2+z^2\right)^2}$$

$$=\frac{\frac{xy+yz+xz}{xyz}+2\left(x^2+y^2+z^2\right)}{\left(x^2+y^2+z^2\right)^2}\le \frac{3\left(x^2+y^2+z^2\right)}{\left(x^2+y^2+z^2\right)^2}=\frac{3}{x^2+y^2+z^2}=\text{R.H.S}$$

Answer 2

The Lev Radzivilovski's proof.

We need to prove that $$\sum_{cyc}\left(\frac{1}{x^2+y^2+z^2}-\frac{1}{x^5+y^2+z^2}\right)\geq0$$ or $$\sum_{cyc}\frac{x^5-x^2}{x^5+y^2+z^2}\geq0$$ or $$\sum_{cyc}\frac{x^2-\frac{1}{x}}{x^2+\frac{y^2+z^2}{x^3}}\geq0.$$ Now, if $x\geq1$ so $$x^2-\frac{1}{x}\geq0$$ and $$x^2+\frac{y^2+z^2}{x^3}\leq x^2+y^2+z^2.$$ If $0<x\leq1$ so $$x^2-\frac{1}{x}\leq0$$ and $$x^2+\frac{y^2+z^2}{x^3}\geq x^2+y^2+z^2.$$ Thus, $$\sum_{cyc}\frac{x^2-\frac{1}{x}}{x^2+\frac{y^2+z^2}{x^3}}\geq\sum_{cyc}\frac{x^2-\frac{1}{x}}{x^2+y^2+z^2}\geq\frac{x^2+y^2+z^2-xy-xz-yz}{x^2+y^2+z^2}\geq0.$$