Inequality involving the length of a vector and its components

Question

Given a vector $\overrightarrow{v}$ of length $|\overrightarrow{v}|$ with components $x,y,z\gt 0$ how to prove the following inequality? $$|\overrightarrow{v}|^2\ge\sqrt 2(xy+yz)$$ Thanks

Answer 1

The inequality is equivalent to $$(x^2+y^2+z^2)^2 \geqslant 2(xy + yz)^2$$

$$\iff y^4 - (4xz) \cdot y^2 + (x^2+z^2)^2 \geqslant 0$$

In terms of $y^2$, this is a quadratic which will remain positive if the discriminant is negative, so all we need is to show $$4(xz)^2 \leqslant (x^2+z^2)^2 \iff (x^2-z^2)^2 \geqslant 0$$