# RMS-AM-GM-HM Inequality

1. Nov 13, 2013

### cheiney

1. The problem statement, all variables and given/known data

I am asked to discover and prove the inequality relationship between root mean square, arithmetic mean, geometric mean, and harmonic mean.

2. Relevant equations
Let a,b, be non-negative integers.
(a-b)2 ≥ 0 and (√a-√b)2 ≥ 0

3. The attempt at a solution

Using (a-b)2 ≥ 0 and (√a-√b)2 ≥ 0, I was able to show that AM ≥ GM , GM ≥ HM, and RMS ≥ GM, but I haven't really been able to show that RMS ≥ AM and I was wondering if someone could point me in the right direction.

I used (a-b)2 ≥ 0 and did some algebra to show that √((a2+b2)/2) ≥ √ab

But I don't know if I can use that to show RMS ≥ AM.

Thanks in advance to anyone who can offer some insight.

2. Nov 13, 2013

### Dick

You know a^2+b^2>=2ab. Here's a hint. Add a^2+b^2 to both sides.

3. Nov 13, 2013

### cheiney

Thanks for the help! It clarified the step I was missing. From there, I would get (a^2+b^2)/2>=((a+b)^2)/4 and then take the square root to get RMS>=AM.

4. Nov 13, 2013

### Dick

You're welcome. Good use of the hint!