How would I prove the following using the AGM inequality?

Question

Question 17. Let $x,y\in\mathbb R$, $x,y\geq0$. Prove that $$(\sqrt x+\sqrt y)^2\geq2\sqrt{2(x+y)\sqrt{xy}}.$$

I believe I have to use AGM multiple times, but I am not exactly sure how

Answer 1

AM-GM for 2 numbers is $\frac{a+b}{2}\geq \sqrt{ab}$. Let $a=x+y$ and $b=2\sqrt{xy}$:

$$\frac{x+y+2\sqrt{xy}}{2}\geq \sqrt{2(x+y)\sqrt{xy}}$$

Multiply both sides by 2 and factor the left hand side to get:

$$x+y+2\sqrt{xy}\geq 2\sqrt{2(x+y)\sqrt{xy}}$$

$$(\sqrt{x}+\sqrt{y})^2\geq 2\sqrt{2(x+y)\sqrt{xy}}$$