Square root of two positive integers less than or equal to the sum of both integers direct proof Please help with this problem.
If x and y positive integers, show:
$$2\sqrt{xy} \le x + y $$

Observe that $(\sqrt{x} - \sqrt{y})^2 = x - 2\sqrt{xy} + y \geq 0$.

 Rationale:  For any real numbers a,b, $(a - b)^2 = a^2 - 2ab + b^2 \geq 0 \implies a^2 + b^2 \geq 2ab$.  Since $x, y > 0$, $\sqrt{x}, \sqrt{y}$ are real numbers.  Thus, if we set $a = \sqrt{x}$ and $b =\sqrt{y}$, we obtain $$(\sqrt{x} - \sqrt{y})^2 = x - 2\sqrt{xy} + y \geq 0 \iff x + y \geq 2\sqrt{xy}$$

