# Proving $\sqrt{x + y} \le \sqrt{x} + \sqrt{y}$

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.

#### Solutions Collecting From Web of "Proving $\sqrt{x + y} \le \sqrt{x} + \sqrt{y}$"

For positive $x, y$:

$$\sqrt{x + y} \leq \sqrt{x} + \sqrt{y} \iff \left(\sqrt{x + y}\right)^2 \leq \left(\sqrt x + \sqrt y\right)^2 \iff \color{green}{\bf x + y \leq x + y }\color{blue}{+ \bf 2\sqrt{xy}}$$

What can you conclude about the leftmost “inequality”, given its equivalence to the ${\bf rightmost\;inequality}$?

Hint: For positive numbers $a$ and $b$,
$$a\leq b\iff a^2\leq b^2.$$

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.

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

$$\sqrt{x+y} < \sqrt{x}+\sqrt{y} \,.$$