Proof that $|x|+|y|\leq\sqrt{2(x^2+y^2)}$ How do I prove that for $x,y\in\mathbb{R}$ we have $|x|+|y|\leq\sqrt{2(x^2+y^2)}$?
I thought that $(|x|+|y|)^2=x^2+y^2+2|x||y|\leq2(x^2+y^2)$, but I'm not sure why that holds.

By the Cauchy-Schwarz inequality, we have:
$$\sqrt{2(x^2+y^2)}=\sqrt{(1+1)\left(|x|^2+|y|^2\right)}\geq|x|+|y|.$$
The CS inequality is the last step, applied with $(|x|,|y|)$ and $(1,1)$. Inequality says the scalar product is at most as great as the product of the norms:
$$|x|+|y|=(|x|,|y|)\cdot(1,1)\leq\|(|x|,|y|)\|\|(1,1)\|=\sqrt{|x|^2+|y|^2}\sqrt{1^2+1^2}=\sqrt{(|x|^2+|y|^2)(1+1)}.$$
