TITLE: How to prove $\sqrt{x + y} \le \sqrt{x} + \sqrt{y}$?
QUESTION [16 upvotes]: 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.

REPLY [4 votes]: Draw a right triangle with the two sides making the $90^\circ$ angle of length $\sqrt{x}$ and $\sqrt{y}$. Then, by the Pytagorean Theorem, the Hypotenuse is $\sqrt{x+y}$.
Since the sum of two edges exceeds the third edge you get 
$$\sqrt{x+y} < \sqrt{x}+\sqrt{y} \,.$$