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

Question

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

Answer 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.

Answer 2

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!

Answer 3

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)

Answer 4

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