Inequality: $x^2+y^2+z^2 \geq \sqrt{2}x(z+y)$ How can I prove the following inequality:
$$x^2+y^2+z^2 \geq \sqrt{2}x(z+y)?$$
Thanks!

We use here that for all $(a,b)\in \mathbb{R}^2$ : 
$$a^2+b^2\ge 2ab.$$ 
Using this with $a=\frac{x}{\sqrt{2}}$ and $y$ gives us 
$$\frac{x^2}{2}+y^2 \ge \sqrt{2}xy.$$ 
Then we use again the first inequality with $a=\frac{x}{\sqrt{2}}$ and $z$ that gives us $$\frac{x^2}{2}+z^2 \ge \sqrt{2}xz.$$ 
Then sum this two inequalities.
