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:
$$2\sqrt{xy} \le x + y $$

It is a standard $AM-GM$ inequality and it can be proved in many ways and here's a one kind of approach:

Result: If $a_1,a_2,\cdots,a_n$ are positive numbers and  $a_1\cdots a_n=1$ then $$a_1+a_2+\cdots+a_n \geq n$$

Take $$a_1=\frac{x}{\sqrt{xy}}$$ and $$a_2=\frac{y}{\sqrt{xy}}$$ and apply above result to see $$\frac{x}{\sqrt{xy}}+\frac{y}{\sqrt{xy}} \geq 2\Rightarrow \frac{x+y}{\sqrt{xy}} \ge 2 \Rightarrow x+y \geq2 \sqrt{xy}$$ 
