Question

If A and G are the A.M and G .M respectively between any two distinct positive numbers a and b, then show that A>G.

Answer 1

let a,b,c be both in A.P and GP

so b=A= $\frac{a+c}{2}$

so, b = G = $\sqrt{ac}$

A-G =$\sqrt{ac} -\frac{a+c}{2}$

by solving you'll get,

A-G = $\frac{(\sqrt{a}- \sqrt{c})^{2}}  {2}$

as the value is squared the value has to be positive

so,

$A-G \geq 0$

$A\geq G$

hence proved