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.
 A: This is just a rearrangement of the famous AM-GM inequality!
It states that $\frac{a+b}{2} \geq \sqrt{ab}$
for $a,b \geq 0$.

The proof is clear from the picture!
Now multiply by $\sqrt{ab}$ on both sides and rearrange with ease to get the desired: $\frac{2ab}{a+b}\leq\sqrt {ab}$.
A: 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.
A: A much simpler way is to use the relation between the harmonic and geometric means:
$$ \frac{2}{\frac{1}{a}+\frac{1}{b}} \leq \sqrt{ab}$$
adding the fraction in the denominator of the left hand side: we have:
$$ \frac{2}{\frac{b}{ab}+\frac{a}{ab}} \leq \sqrt{ab}$$
And the relation you listed follows.
See also https://en.wikipedia.org/wiki/Mean#Relationship_between_AM.2C_GM.2C_and_HM
