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.

So what you're trying to prove is
$$
x^2+y^2\ge 2|x||y|
$$
which is the famous AM-GM inequality. You can prove it by completing the square.
$$
x^2+y^2-2|x||y|=(|x|-|y|)^2\ge0
$$   
