# A.M.>G.M. of four numbers

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## Solution 1

Hint: If you can use the AM-GM on 2 numbers, then you should use that.

First prove with this inequality on $(a,b)$ $$\frac{a+b}{2} \geq \sqrt{ab}$$

Then prove with this inequality on $(c,d)$ $$\frac{c+d}{2} \geq \sqrt{cd}$$

Then use the AM-GM on 2 numbers again on the terms above.

## Solution 2

The inequality $\text{AM}>\text{GM}$ is equivalent to $\log\text{AM}>\log\text{GM}$, because the logarithm is a monotone increasing function.

Expanding $\log\text{AM}$ and $\log\text{GM}$,

\begin{align} \log\text{AM} & = \log \frac{a+b+c+d}{4} \\ \log\text{GM} & = \log(abcd)^{1/4} = \frac{\log a+\log b+\log c+\log d}{4} \end{align}

The result then follows because the logarithm is a concave function (see Jensen's inequality)

## Solution 3

we have by AM-GM: $$\frac{\frac{a+b}{2}+\frac{c+d}{2}}{2}\geq \sqrt{\frac{a+b}{2}\frac{c+d}{2}}\geq \sqrt{\sqrt{ab}\sqrt{cd}}$$

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### Wisha

Updated on February 07, 2020

• Wisha over 3 years

Prove that arithmetic mean of $4$ numbers is greater than geometric mean of the same $4$ numbers, i.e. prove that $$\dfrac{a+b+c+d}{4} > (abcd)^{\frac1{4}}$$

• 5xum almost 8 years
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it?
• Ruben almost 8 years
For correctness, I think you should add that a, b, c, d should all be different and non-negative.
• Thomas Ahle almost 8 years
I don't think your expansion is correct. You need to divide everything by $4^4$.
• NeilRoy almost 8 years
• wythagoras almost 8 years
This is true by Jensen's inequality.
• Luis Mendo almost 8 years
@wythagoras Thanks. I've included that
• Ruben almost 8 years
$4^4$ > 24, making the proof incorrect. Would have been a nice proof though!
• wythagoras almost 8 years
Well you could make all kinds of horrible estimates (e.g $a^4+b^4+c^4+d^4>4abcd$ by the rearrangement inequality) but I wouldn't recommend that.
• Wisha almost 8 years
@MonKeePoo yeh thats right... but if we shift $4^4=256$ to rhs...the lhs will still be huge!
Why is this clearly bigger than $abcd$? It could be the case that for example $4c^3d<0$.