Proof of Inequality 1: $0\le \(\|x\|-\left|y\right|{\)}^2$ = $x^2\+y^2-2\left|x y\right|$ ⇒ $\left|x y\right|$ $\le \left(x^2\+y^2\right)\/2$.

Proof of Inequality 2: $0\le \(\left|x\right|\+\|y\|{\)}^2$ = $x^2\+y^2\+2\left|x y\right|$ $\le x^2\+y^2\+2\left(x^2\+y^2\right)\/2\=2\left(x^2\+y^2\right)$.

Since $\(\left|x\right|\+\|y\|{\)}^2\le 2\left(x^2\+y^2\right)$, it follows that $\left|x\right|\+\|y\|\le \sqrt{2} \sqrt{x^2\+y^2}$.

(Note the use of Inequality 1 in the proof of Inequality 2.)