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Prove that for all real numbers $x,y$ and $z$ that $x^2+y^2+z^2 \geq x\sqrt{y^2+z^2}+y\sqrt{x^2+z^2}$.

It looks like AM-GM should be used here but the square roots make it difficult. So maybe Cauchy-Schwarz works?

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This is simply 2 applications of $ab\leq\frac{1}{2}(a^2+b^2)$: $$ x\sqrt{y^2+z^2}\leq\frac{1}{2}(x^2+y^2+z^2),\quad y\sqrt{x^2+z^2}\leq\frac{1}{2}(y^2+x^2+z^2) $$ Now, sum the above 2 inequalities.

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Note that (by AM-GM),

$\frac{x^2+(y^2+z^2)}{2}\geq\sqrt{x^2(y^2+z^2)}=x\sqrt{y^2+z^2}$.

and similarly,

$\frac{y^2+(x^2+z^2)}{2}\geq\sqrt{y^2(x^2+z^2)}=y\sqrt{x^2+z^2}$.

Summing them up gives:

$x^2+y^2+z^2\geq x\sqrt{y^2+z^2}+y\sqrt{x^2+z^2}$

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