How can I prove the inequality relationship between RMS, AM, GM, and HM?

[cheiney]

Homework Statement

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

Homework Equations

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

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.

 

[Dick]

Homework Statement

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

Homework Equations

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

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.

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

 

[cheiney]

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

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.

 

[Dick]

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.

You're welcome. Good use of the hint!