<s>[INST] Prove that $\frac{x^2+y^2}{z}+\frac{y^2+z^2}{x}+\frac{z^2+x^2}{y}\geq 2(x+y+z)$ $x,y,z \in \mathbb R^+$ different than $0$, Prove that :
$$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$
My attempt:
First, we should prove that :
$$2\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x} \right)\geq 2(x+y+z)$$
I will consider it as proven, we gonna start by :
$$(x-y)^2 \geq 0 \iff x^2+y^2 \geq 2xy $$
by the same method we have :
$x^2+z^2 \geq 2xz$ and $y^2+z^2 \geq 2yz$
so we can replace $2xy$ by $x^2+y^2$ in this inequality
$$2\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x} \right)\geq 2(x+y+z)$$
so :
$$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$
I'm new to AM-GM inequality, if it exists a better proof using it please post it.
 [/INST] $$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$
$$\Leftrightarrow (x^2 + y^2 + z^2) \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \geq 3(x+y+z)$$
Note that
$$(x^2 + y^2 + z^2) \geq \frac{1}{3} (x+y+z)^2 $$ (Cauchy-Schwarz ineq., equality: x = y = z)
and
$$(x+y+z) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \geq (1+1+1)^2 = 9$$ (Cauchy-Schwarz ineq., equality: $x=y=z$).
Applying them implies the original inequality (equality holds when $x=y=z$).
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