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Prove

$$ \frac{1}{x^5+y^2+z^2} + \frac{1}{x^2+y^5+z^2} + \frac{1}{x^2+y^2+z^5} \leq \frac{3}{x^2+y^2+z^2} ,$$

when $xyz \geq 1$ ($x,y,z$ are positive real numbers).

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

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closed as off-topic by Shaun, Thomas Shelby, Aqua, user21820, mrtaurho Mar 19 at 6:38

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By Cauchy-Schwarz's inequality:

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

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

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

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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, $$\sum_{cyc}\frac{x^2-\frac{1}{x}}{x^2+\frac{y^2+z^2}{x^3}}\geq\sum_{cyc}\frac{x^2-\frac{1}{x}}{x^2+y^2+z^2}\geq\frac{x^2+y^2+z^2-xy-xz-yz}{x^2+y^2+z^2}\geq0.$$

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