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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 $$\dfrac{a+b+c+d}{4} > (abcd)^{\frac1{4}}$$


marked as duplicate by user147263, Joel Reyes Noche, Ali Caglayan, David K, Did Jun 13 '15 at 15:47

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

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.


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)

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    $\begingroup$ This is true by Jensen's inequality. $\endgroup$ – wythagoras Jun 10 '15 at 14:24
  • $\begingroup$ @wythagoras Thanks. I've included that $\endgroup$ – Luis Mendo Jun 10 '15 at 14:26
  • $\begingroup$ hi your method is interesing. $\endgroup$ – Wisha Jun 10 '15 at 14:30
  • $\begingroup$ can u pls tell me what is concave function? $\endgroup$ – Wisha Jun 10 '15 at 14:31
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    $\begingroup$ It's a function that lies above any of its chords. It's the opposite of a convex function. Did you follow the link in my question? $\endgroup$ – Luis Mendo Jun 10 '15 at 14:32

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


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!

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    $\begingroup$ I don't think your expansion is correct. You need to divide everything by $4^4$. $\endgroup$ – Thomas Ahle Jun 10 '15 at 14:18
  • $\begingroup$ Sorry ... my bad! $\endgroup$ – NeilRoy Jun 10 '15 at 14:24
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    $\begingroup$ $ 4^4 $ > 24, making the proof incorrect. Would have been a nice proof though! $\endgroup$ – Ruben Jun 10 '15 at 14:28
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    $\begingroup$ 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. $\endgroup$ – wythagoras Jun 10 '15 at 14:30
  • $\begingroup$ @MonKeePoo yeh thats right... but if we shift $4^4=256$ to rhs...the lhs will still be huge! $\endgroup$ – NeilRoy Jun 10 '15 at 14:32

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