Question

Prove that for two positive numbers A.M.( Arithmetic mean) $\ge$ G.M. (Geometric mean).

Answer 1

Let $a,b>0$ be two positive numbers.

Now A.M. and G.M. of $a,b$ are $\frac{a+b}{2}$ and $\sqrt{ab}$ respectively.

We have,

$(\sqrt{a}-\sqrt{b}{)}^2\ge 0$

or, $a+b-2\sqrt{ab}\ge 0$

or, $\frac{a+b}{2}\ge \sqrt{ab}$

or, A.M. $\ge$ G.M.