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

Harsh Kumar
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Wisha
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    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? – 5xum Jun 10 '15 at 14:12
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    For correctness, I think you should add that a, b, c, d should all be different and non-negative. – Ruben Jun 10 '15 at 14:21

4 Answers4

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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.

wythagoras
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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)

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

Dr. Sonnhard Graubner
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A very crude proof:

We have the equation in L.H.S. as: $\dfrac{a+b+c+d}{4}$ and R.H.S. as $(abcd)^{1/4}$

Now shifting the $\dfrac{1}{4}^{th}$ power from r.h.s. to l.h.s., we get:

L.H.S. as $\bigg[\dfrac{a+b+c+d}{4}\bigg]^4$

$=\dfrac{a^4+4 a^3 b+4 a^3 c+4 a^3 d+6 a^2 b^2+12 a^2 b c+12 a^2 b d+6 a^2 c^2+12 a^2 c d+6 a^2 d^2+4 a b^3+12 a b^2 c+12 a b^2 d+12 a b c^2+24 a b c d+12 a b d^2+4 a c^3+12 a c^2 d+12 a c d^2+4 a d^3+b^4+4 b^3 c+4 b^3 d+6 b^2 c^2+12 b^2 c d+6 b^2 d^2+4 b c^3+12 b c^2 d+12 b c d^2+4 b d^3+c^4+4 c^3 d+6 c^2 d^2+4 c d^3+d^4 \color{blue}{+ 24abcd}}{4^4}$

which is clearly bigger than $abcd$ in the R.H.S.

Hope this satisfies!

NeilRoy
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