Square root of two positive integers less than or equal to the sum of both integers direct proof

Question

Please help with this problem.

If x and y positive integers, show: $$2\sqrt{xy} \le x + y $$

Answer 1

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 $$\frac{x}{\sqrt{xy}}+\frac{y}{\sqrt{xy}} \geq 2\Rightarrow \frac{x+y}{\sqrt{xy}} \ge 2 \Rightarrow x+y \geq2 \sqrt{xy}$$

Answer 2

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

Answer 3

Since $x$ and $y$ are positive.

Your question is equivalent to $G.M\le A.M$.

$GM$ means geometric mean and $AM$ means arithmetic mean of $x&y$.