<s>[INST] How to prove $\sqrt{x + y} \le \sqrt{x} + \sqrt{y}$? How do I prove $\sqrt{x + y} \le \sqrt{x} + \sqrt{y}$? for $x, y$ positive?
This should be easy, but I'm not seeing how.  A hint would be appreciated.
 [/INST] Putting everything together, suppose $x,y>0$. Then $0\le2\sqrt{xy}$. Hence:
$$ 
\sqrt{x+y} = \sqrt{x+0+y}
\le \sqrt{x+2\sqrt{xy}+y} 
= \sqrt{(\sqrt{x}+\sqrt{y})^2} 
= \sqrt{x}+\sqrt{y} 
$$
as desired. Note that this relied on the fact that $f(x)=\sqrt{x}$ is monotonically increasing.
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