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$a$ 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.

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    $\begingroup$ This cannon be correct since you have equality if $a=b$. $\endgroup$ – A.S. Jan 31 '16 at 5:48
  • $\begingroup$ It's mostly correct (except you should write it backwards and replace < with <=) but a little complicated. It is simpler to square each side, simplify by $ab$ and get rid of denominators $\endgroup$ – Ewan Delanoy Jan 31 '16 at 5:49
  • $\begingroup$ I tried this approach and got 0 $<=$ $(a-b)^2$, which could be $0$ if $a$ and $b$ are equal, or positive in any other case, which makes the inequality true. Hopefully I did the math right. $\endgroup$ – Chris Jan 31 '16 at 6:03
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This is just a rearrangement of the famous AM-GM inequality!

It states that $\frac{a+b}{2} \geq \sqrt{ab}$

for $a,b \geq 0$.

enter image description here

The proof is clear from the picture!

Now multiply by $\sqrt{ab}$ on both sides and rearrange with ease to get the desired: $\frac{2ab}{a+b}\leq\sqrt {ab}$.

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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

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