# Using the AM-GM inequality on 2 elements to deduce it's true for 4?

I was asked a two part question, first prove the AM-GM inequality for n=2 (with some extra)
$\frac{2}{\frac{1}{a}+\frac{1}{b}}\leq\sqrt{ab}\leq\frac{a+b}{2}$
I was able to do that without too many problems.

Then I'm asked to deduce from the above that
$\frac{4}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}\leq\sqrt[4]{abcd}\leq\frac{a+b+c+d}{4}$
I think it just have something to do with applying the inequity proven earlier a few times, it just feels like a trick question where I'm missing something. Any ideas?

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$$\frac{a+b+c+d}{4} = \frac{\frac{a+b}{2} + \frac{c+d}{2}}{2}$$ Also, the "extra" is called the harmonic mean. – Arthur Nov 3 '12 at 13:39

## 1 Answer

Here is a big hint:

$a+b\ge 2\sqrt {ab}$ and $c+d\ge 2\sqrt{cd}$.Now apply the AM-GM inequality again.

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Got it, thanks... – Nescio Nov 3 '12 at 13:41