<s>[INST] $ x \ge 0\text{ and } y \ge 0 \implies \frac{x+y}{2} \ge \sqrt{xy} $ The above applies $\forall x,y \in \mathbb{R}$
I've tried: $x + y \ge 0$
$$x + y \ge x$$
$$ (x + y)^2 \ge 2xy$$
$$\frac{(x + y)^2}{2} \ge xy$$
But the closest I get is $\dfrac{x+y}{\sqrt{2}} \ge \sqrt{xy}$
Any ideas?
 [/INST] $$(x-y)^2 \ge 0$$
$$x^2 - 2xy + y^2 \ge 0 $$
$$x^2 + y^2 \ge 2xy $$
$$x^2 + 2xy + y^2 \ge 4xy $$
$$(x+y)^2 \ge 4xy  $$
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