Someone's Inequality ? $(a + b)^2 \le 2(a^2 + b^2) $ 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.

This inequality can as weel be seen as a particular case of the equivalence between $p$-norms in $\Bbb R^n$. Indeed for $1<p\leq q<\infty$, it holds
$$\|x\|_q \leq \|x\|_p \leq n^{1/p-1/q}\|x\|_q$$
In the particular case $n=2$, $p=1$ and $q=2$ we get
$$(|a|+|b|)=\|(a,b)\|_1\leq 2^{1-1/2}\|(a,b)\|_2 =\sqrt{2(a^2+b^2)},$$
which is even slightly tighter as $|a+b| \leq |a|+|b|$.
I should nevertheless point out that the equivalence between $p$-norms is proved using the Cauchy-Schwarz (or more generally the Hölder) inequality mentioned by @Ihf
