For real $a, b$ then $(a + b)^2 \le 2(a^2 + b^2) $
This fairly trivial inequality crops up a lot in my reading on (Lebesgue) integration, is it named after someone ? It extends rather obviously for positive reals to $a^2 + b^2 \le (a + b)^2 \le 2(a^2 + b^2) $.
Proof (if you need it):
$0 \le (a - b)^2 = a^2 + b^2 -2ab \implies 2ab \le a^2 + b^2$
$(a + b)^2 = a^2 + b^2 + 2ab$ which by previous $ \le 2(a^2 + b^2) $.
Application:
If $f, g$ are positive functions then $(f + g)^2$ is integrable $\iff$ $f^2, g^2$ are integrable since $f^2 + g^2 \le (f + g)^2 \le 2(f^2 + g^2) $ pointwise.
That inequality probably does not have a name as it is so basic. In any case it can be viewed as a special case of Young's inequality.