<s>[INST] Prove $\frac{2ab}{a+b}\leq\sqrt {ab}$ $a$ and $b$ are both positive real numbers. I'm supposed to work backwards (i.e. start with what I'm trying to prove and change it until something is absolutely true, then start from what is absolutely true in my proof).
Here's my attempt:
$\frac{2ab}{a+b}\leq\sqrt {ab}$
$\frac{2a^2b^2}{(a+b)^2}\leq{ab}$
$\frac{2a^2b^2}{a^2+2ab+b^2} - ab \leq 0$
$\frac{2a^2b^2-ab(a^2+2ab+b^2)}{a^2+2ab+b^2}\leq 0$
$\frac{2a^2b^2-a^3b-2a^2b^2-ab^3}{a^2+2ab+b^2}\leq 0$
$\frac{-a^3b-ab^3}{a^2+2ab+b^2}\leq0$
Because $a$ and $b$ are positive, this guarantees the left side is negative, making the inequality true.
 [/INST] Let $a,b\in\mathbb{R}$ with $a,b>0$.
$$\begin{align*}\frac{2ab}{a+b}\leq \sqrt{ab}&\Longleftrightarrow \frac{2}{a+b}\leq\frac{\sqrt{ab}}{ab}\\&\Longleftrightarrow \frac{2}{a+b}\leq\frac{1}{\sqrt{ab}}\\&\Longleftrightarrow\sqrt{ab}\leq\frac{a+b}{2}\\&\Longleftrightarrow 2\sqrt{ab}\leq a+b\\&\Longleftrightarrow 0\leq a-2\sqrt{ab}+b\\&\Longleftrightarrow 0\leq (\sqrt{a}-\sqrt{b})^2\end{align*}$$ where the last inequality is always true.
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