Why is this inequality true? $(a+b)^2\leq 2(a^2+b^2)$ Why is this inequality true? $a,b$ are real numbers.
$$
(a+b)^2=a^2+2ab+b^2\leq 2(a^2+b^2)
$$
I know $(a+b)^2=a^2+2ab+b^2 \geq 0$, but then?

Equivalently: $$a^2+2ab+b^2 \leq 2a^2+2b^2$$
Equivalently: $$0 \leq a^2-2ab+b^2$$
Equivalently: $$0 \leq (a-b)^2$$
This latter expression is true: squares are nonnegative.
