<s>[INST] A quick way to prove the inequality $\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$ Can anyone suggest a quick way to prove this inequality?
$$\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$$
 [/INST] assuming $$x\geq0 \land y\geq0$$
$$\dfrac{\sqrt{x}+\sqrt{y}}{2} \leq \sqrt{\dfrac{x+y}{2}}$$
Squaring:
$$\dfrac{x+y+2\sqrt{xy}}{4} \leq \dfrac{x+y}{2} \implies \dfrac{\sqrt{xy}}{2} \leq \dfrac{x+y}{4}$$
Squaring again
$$\dfrac{\sqrt{xy}}{2} \leq \dfrac{x+y}{4} \implies \dfrac{xy}{4} \leq \dfrac{x^2+2xy+y^2}{16}\implies$$
$$0 \leq \dfrac{x^2-2xy+y^2}{16} \implies 0 \leq \dfrac{(x-y)^2}{4}$$
But we now that $a^2\geq 0 \forall a \in \mathbb{R}$ so it's true...
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