Inequality: $x^2+y^2+z^2 \geq \sqrt{2}x(z+y)$

Question

How can I prove the following inequality:

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

Thanks!

Answer 1

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.

Answer 2

Obviously, is enough to prove for $x\ge0$, $y\ge0$, $z\ge0$. By homogeneity, can be supposed wlog that $x^2+y^2+z^2=1$ Using Lagrange multipliers with the problem $$f(x,y,z)=\sqrt{2}x(z+y),\qquad x^2+y^2+z^2=1,$$ we have: $$\sqrt{2}(z+y)=2\lambda x,$$ $$\sqrt{2}x=2\lambda y,$$ $$\sqrt{2}x=2\lambda z,$$ with nontrivial solutions for $\lambda=0,\pm1$...