How to prove $\sqrt{x + y} \le \sqrt{x} + \sqrt{y}$?

For positive $x, y$, we have:

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

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

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

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

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