Question

If A,G,H be respectively the A.M., G.M. and H.M. between two given quantities a and b, then prove that
A > G > H

Answer 1

Let there are two quantities a and b,

a, b>0

$h=\sqrt{ab}$

$H=\frac{2ab}{a+b}$ $A=\frac{a+b}{2}$

$H\times A=\frac{2ab}{a+b}\times \frac{a+b}{2}=ab=(\sqrt{ab}{)}^2=(G.M.{)}^2$

These mean they are in geometric progression.

All three (A, G, H) follow the rule of geometric progression.

$G=\sqrt{A}\times H$

Now let difference between A and G

$A-G=\frac{a+b}{2}-\sqrt{ab}=(\sqrt{a}{)}^2+({\sqrt{b}}^2)-\frac{2\sqrt{a}\sqrt{b}}{2}$

A>G -----------(1)

Similarly, $G-H=\sqrt{ab}-\frac{2ab}{a+b}=\frac{\sqrt{ab}}{a+b}(\sqrt{a}-\sqrt{b}{)}^2>0$

So, G>H ---------(2)

So, from (1) and (2), we get,

A>G>H