MHB Which mean is larger when using algebraic expressions, A. M. or R. M. S.?

AI Thread Summary
The discussion revolves around comparing the root mean square (R.M.S.) and arithmetic mean (A.M.) for two positive numbers and algebraic expressions. It is established that for specific values, R.M.S. is greater than A.M. The participants explore whether this relationship holds for algebraic expressions, specifically when a = x and b = 1/x. A proof is developed showing that R.M.S. is always greater than or equal to A.M., culminating in the conclusion that (a - b)² ≥ 0 confirms the inequality. The conversation also touches on the importance of sharing questions in the forum rather than through private messages.
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Given two positive numbers a and b, we define the root mean square as follows:

R. M. S. = sqrt{(a^2 + b^2)/2}

The arithmetic mean is given by (a + b)/2.

Given a = 1 and b = 2, which is larger, A. M. or R. M. S. ?

A. M. = sqrt{1•2}

A. M. = sqrt{2}

R. M. S. = sqrt{(1^2 + 2^2)/2}

R. M. S. = sqrt{5/2}

Conclusion:

R. M. S. > A. M.

Question:

Can the same process be done if a and b represent two algebraic expressions?

Say a = x and b = 1/x, which is larger, A. M. or R. M. S. ?
 
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Can you prove that in general RMS ≥ AM? When does equality occur?
 
R. M. S. = sqrt{(a^2 + b^2)/2} > or = (a + b)/2.

Should I begin by squaring both sides?
 
RTCNTC said:
R. M. S. = sqrt{(a^2 + b^2)/2} > or = (a + b)/2.

Should I begin by squaring both sides?

Yes, that's how I'd begin to get rid of the radical...:D
 
I will play with this question on my next day off. Thank you very much.
 
[sqrt{(a^2 + b^2)/2}]^2 ≥ [(a + b)/2]^2.

(a^2 + b^2)/2 ≥ (a + b)^2/4

4 • (a^2 + b^2)/2 ≥ (a + b)^2/4 • 4

2(a^2 + b^2) ≥ (a + b)^2

2a^2 + 2b^2 ≥ (a^2 + 2ab + b^2)

2a^2 - a^2 + 2b^2 - b^2 - 2ab ≥ 0

a^2 + b^2 - 2ab ≥ 0

Where do I go from here?
 
RTCNTC said:
[sqrt{(a^2 + b^2)/2}]^2 ≥ [(a + b)/2]^2.

(a^2 + b^2)/2 ≥ (a + b)^2/4

4 • (a^2 + b^2)/2 ≥ (a + b)^2/4 • 4

2(a^2 + b^2) ≥ (a + b)^2

2a^2 + 2b^2 ≥ (a^2 + 2ab + b^2)

2a^2 - a^2 + 2b^2 - b^2 - 2ab ≥ 0

a^2 + b^2 - 2ab ≥ 0

Where do I go from here?

Try factoring the LHS...:D
 
The LHS becomes (a - b)^2 ≥ 0.

Do I proceed by taking the square root on both sides?
 
RTCNTC said:
The LHS becomes (a - b)^2 ≥ 0.

Do I proceed by taking the square root on both sides?

You could, but it's not necessary. If you do, recall:

$$\sqrt{x^2}=|x|$$
 
  • #10
Am I done with the prove?
 
  • #11
RTCNTC said:
Am I done with the prove?

You have:

$$(a-b)^2\ge0$$

Given that a - b is a real number, this must be true, and so we are done with the proof. :D
 
  • #12
Great. I will return to the quadratic inequality questions tomorrow as we travel through the David Cohen book. It is truly one of the most challenging precalculus books out there. I will text 5 questions through PM in 15 minutes.
 
  • #13
RTCNTC said:
Great. I will return to the quadratic inequality questions tomorrow as we travel through the David Cohen book. It is truly one of the most challenging precalculus books out there. I will text 5 questions through PM in 15 minutes.

I would rather you post the questions in the forums rather than sending them by PM, as per our rules. :D
 
  • #14
The questions are not math questions. Check your PM.
 
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