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Prove that if $x,y,z$ are positive real numbers then the following inequality holds $$ \frac {x+y}{x^2+y^2} + \frac {y+z}{y^2+z^2} + \frac {z+x}{z^2+x^2} \leq \frac 1x + \frac 1y + \frac 1z . $$

I have tried everything to listed in the title like the AM-GM inequality and the Cauchy-Schwarz Inequality but it seems to be getting me nowhere. I also tried brute force but that really just led me to going in circles.

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We have $a^2+b^2\ge 2ab$ for all $a,b\in\mathbb R$, because this is equivalent to $(a-b)^2\ge 0$, which is true.

Since $x,y,z>0$, we get:

$$\frac {x+y}{x^2+y^2} + \frac {y+z}{y^2+z^2} + \frac {z+x}{z^2+x^2}$$

$$\leq \frac{x+y}{2xy}+\frac{y+z}{2yz}+\frac{z+x}{2zx}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$

Equality holds if and only if $x=y=z$.

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  • $\begingroup$ All too easy. +1 $\endgroup$ – Mark Viola Mar 9 '16 at 2:32

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