<s>[INST] Subadditivity of square root function Any hint to prove 
$$ \sqrt{x+y} \le \sqrt{x} + \sqrt{y}, \qquad \forall x,y \ge 0 $$
 [/INST] I'll try the hard way
and use calculus.
Let
$f(y)
= \sqrt{x+y} - \sqrt{x} - \sqrt{y}
$.
$f(0)
=0
$.
$f'(y)
=\frac1{2\sqrt{x+y}}-\frac1{2\sqrt{y}}
$.
Since
$x+y \ge y$,
$\sqrt{x+y} \ge \sqrt{y}$
so
$\frac1{2\sqrt{x+y}}
\le \frac1{2\sqrt{y}}
$.
Therefore
$f'(y) \le 0$
for all $y$.
Since
$f(0) = 0$,
$f(y)
\le 0
$ for all $y$.
Note that this
can be used to show that
if
$g(y)$
is increasing and
$g'(y)$
is decreasing
then
$g(x+y)
\le g(x)+g(y)
$.
Similarly,
if $g'(y)$
is increasing,
$g(x+y)
\ge g(x)+g(y)
$.
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