<s>[INST] 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.
 [/INST] This follows from the Cauchy–Schwarz inequality
$$
|\langle \mathbf{u},\mathbf{v}\rangle| ^2 \leq \langle \mathbf{u},\mathbf{u}\rangle \cdot \langle \mathbf{v},\mathbf{v}\rangle,
$$
with
$
\mathbf{u} = (1,1)$
and
$
\mathbf{v} = (a,b)
$
in $\mathbb R^2$.
</s>