Subadditivity of square root function Any hint to prove 
$$ \sqrt{x+y} \le \sqrt{x} + \sqrt{y}, \qquad \forall x,y \ge 0 $$
 A: Notice that $$\sqrt{x+y}\le \sqrt{x}+\sqrt{y}\Leftrightarrow x+y\le x+y+2\sqrt{xy}.$$
Which clearly holds as $x,y\ge{0}$, and the function $x\rightarrow x^2$ is monotonic and equality occurs when one of $x$ or $y$ is $0$.
A: Another:
$x,y \geq 0$ then the product $\sqrt{x} \sqrt{y}$ have sence and now
$$\sqrt{x + y} \leq \sqrt{x + 2\sqrt{x}\sqrt{y} + y} = \sqrt{(\sqrt{x} + \sqrt{y})^2} = \sqrt{x} + \sqrt{y}$$
A: 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)
$.
A: The inequality is homogeneous, hence we may assume $y=1$ without loss of generality. 
So we just have to prove that $\sqrt{x+1}-\sqrt{x}\leq 1$, pretty easy:
$$ \sqrt{x+1}-\sqrt{x} = \frac{(x+1)-x}{\sqrt{x+1}+\sqrt{x}} = \frac{1}{\sqrt{x+1}+\sqrt{x}}\leq 1.$$
