How can I Prove $\frac{2xy}{x+y}\leq \sqrt{xy}\leq \frac{x+y}{2}$

inequality means

$\sqrt{xy} \le \dfrac{x + y}{2}$ is just the AM–GM inequality. $\dfrac{2xy}{x + y} \le \sqrt{xy}$ is also the AM-GM inequality applied to $1/x$, $1/y$.


For example

$$\frac{2xy}{x+y}\le\sqrt{xy}\iff 4x^2y^2\le xy(x^2+2xy+y^2)\iff$$

$$ \iff xy(x^2+2xy+y^2-4xy)\ge 0\iff xy(x-y)^2\ge 0$$

And since the last rightmost inequality is obvious we're done.

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