<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] Both sides are positive so you may square the inequality ot obtain $$\begin{align*}\left(\frac{\sqrt{x}+\sqrt{y}}{2}\right)^2\le\left(\sqrt{\frac{x+y}{2}}\right)^2 &\iff \frac{x+2\sqrt{xy}+y}{4}\le \frac{x+y}{2} \\&\iff \frac{2\sqrt{xy}}{4}\le\frac{x+y}{2}-\frac{x+y}{4} \\&\iff \sqrt{xy}\le \frac{x+y}{2}\\[0.2cm]& \iff (\sqrt{x}-\sqrt{y})^2\ge 0\end{align*}$$ which is true.
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