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

[mathdad]

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. ?

 

[MarkFL]

Can you prove that in general RMS ≥ AM? When does equality occur?

 

[mathdad]

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

Should I begin by squaring both sides?

 

[MarkFL]

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

 

[mathdad]

I will play with this question on my next day off. Thank you very much.

 

[mathdad]

[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?

 

[MarkFL]

[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

 

[mathdad]

The LHS becomes (a - b)^2 ≥ 0.

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

 

[MarkFL]

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:

$$\sqrt{x^2}=|x|$$

 

[mathdad]

Am I done with the prove?

 

[MarkFL]

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

 

[mathdad]

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.

 

[MarkFL]

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

 

[mathdad]

The questions are not math questions. Check your PM.