prove that AM > GM> HM

Dear student

A . M = \frac{(a + b)}{2} G . M = \sqrt{a . b} H . M = \frac{2 ab}{(a + b)} So, A . M - G . M = \frac{(a + b)}{2} - \sqrt{a . b} = \frac{a + b - 2 \sqrt{a . b}}{2} = \frac{{(\sqrt{a - b})}^{2}}{2} \geq 0 [square of anything is always positive] G . M - H . M = \sqrt{a . b} - \frac{2 ab}{(a + b)} = \sqrt{a . b} [\frac{(a + b - 2 \sqrt{ab})}{a + b}] = \frac{\sqrt{a . b} {(\sqrt{a} + \sqrt{b})}^{2}}{a + b} \geq 0 [square of anything is always positive] So proved that A . M \geq G . M \geq H . M

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