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.
 A: The claim immediately follows from
$$\bigl(|x|+|y|\bigr)^2\leq\bigl(|x|+|y|\bigr)^2+\bigl(|x|-|y|\bigr)^2=2\bigl(|x|^2+|y|^2\bigr)\ .$$
A: 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.
$$
x^2+y^2-2|x||y|=(|x|-|y|)^2\ge0
$$   
A: 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:
$$|x|+|y|=(|x|,|y|)\cdot(1,1)\leq\|(|x|,|y|)\|\|(1,1)\|=\sqrt{|x|^2+|y|^2}\sqrt{1^2+1^2}=\sqrt{(|x|^2+|y|^2)(1+1)}.$$
A: The key is recognizing that $|x|^2 = x^2 \,\,\,\, \forall x\in\mathbb{R}$, and that the inequality you start with can simply be squared since both sides are guaranteed to be positive:
$$|x| + |y| \leq \sqrt{2(x^2+y^2)}$$
$$\Leftrightarrow x^2 + 2 |x| |y| + y^2 \leq 2x^2 + 2y^2$$
The rest is rearrangement and factoring as shown by others.
