Question

How


a.m≥g.m≥h.m. prove without using numbers. ?

Answer 1

We need a little correction to the question.
It should be AM > GM > HM.

Proof.

Let us take two unequal positive numbers, whose arithmetic mean is AM, geometric mean is GM and harmonic mean is HM.

∴ AM = (a + b)/2,

GM = √(ab) and

HM = 2ab/(a + b)

We see that, AM × HM

= (a + b)/2 × 2ab/(a + b)

= ab

= GM²

⇒ AM × HM = GM²

Again, AM - GM

= (a + b)/2 - √(ab)

= {a + b - 2√(ab)}/2

= 1/2 (√a - √b)²

⇒ AM > GM, since (√a - √b)² > 0 when a and b are positive numbers and a and b aren't equal

Since, AM × HM = GM² and AM > GM

∴ HM < GM

So, AM > GM > HM [Proved]

I hope it helps you.