Can the geometric and arithmetic means be applied to algebraic expressions?

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Discussion Overview

The discussion revolves around the comparison of the geometric mean (G.M.) and arithmetic mean (A.M.) for both numerical values and algebraic expressions. Participants explore whether the same principles apply when the variables represent algebraic expressions, specifically examining the case where a = x and b = 1/x.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants define the geometric mean and arithmetic mean for positive numbers and calculate them for a = 1 and b = 2, concluding that G.M. > A.M.
  • Participants question whether the same comparison can be made when a and b are algebraic expressions, specifically a = x and b = 1/x.
  • There is a proposal to prove that A.M. is greater than or equal to G.M. in general, with some participants suggesting to start by squaring both sides of the inequality.
  • Participants engage in mathematical manipulation to show the relationship between A.M. and G.M., including factoring and recognizing the square of a binomial.
  • One participant expresses uncertainty about the next steps in the proof, asking for guidance on how to proceed with the factoring process.
  • Another participant identifies that the expression can be recognized as the square of a binomial, leading to the conclusion that the inequality holds for valid values of a and b.

Areas of Agreement / Disagreement

Participants generally agree on the definitions and calculations of G.M. and A.M. for specific numbers, but there is ongoing exploration and debate regarding their application to algebraic expressions. The discussion remains unresolved regarding the broader implications and proofs of the inequalities.

Contextual Notes

Some participants express uncertainty about the steps in proving the inequality, and there are unresolved mathematical steps related to the factoring and interpretation of the expressions involved.

mathdad
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Given two positive numbers a and b, we define the geometric mean and the arithmetic mean as follows

G. M. = sqrt{ab}

A. M. = (a + b)/2

If a = 1 and b = 2, which is larger, G. M. or A. M. ?

G. M. = sqrt{1•2}

G. M. = sqrt{2}

A. M. = (1 + 2)/2

A. M = 3/2

Conclusion: G. M. > A. M.

Correct?

Question:

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

Say a = x and b = 1/x. Which is bigger, G. M. or A. M. ?
 
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RTCNTC said:
Given two positive numbers a and b, we define the geometric mean and the arithmetic mean as follows

G. M. = sqrt{ab}

A. M. = (a + b)/2

If a = 1 and b = 2, which is larger, G. M. or A. M. ?

G. M. = sqrt{1•2}

G. M. = sqrt{2}

A. M. = (1 + 2)/2

A. M = 3/2

Conclusion: G. M. > A. M.

Correct?

Question:

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

Say a = x and b = 1/x. Which is bigger, G. M. or A. M. ?

Note that:

$$\sqrt{2}<\frac{3}{2}$$

Can you prove that in general AM ≥ GM? When does equality occur?
 
I meant to type A. M. > G. M., of course.

To prove that A. M. > G. M. in general, we set A. M. is greater than or equal to sqrt{ab}.

(a + b)/2 > or = sqrt{ab}

Can you get me started?
 
RTCNTC said:
I meant to type A. M. > G. M., of course.

To prove that A. M. > G. M. in general, we set A. M. is greater than or equal to sqrt{ab}.

(a + b)/2 > or = sqrt{ab}

Can you get me started?

Since a and b both represent positive numbers, I would begin by squaring both sides...:D
 
Are you saying to treat it like a radical equation?
 
[(a + b)/2]^2 ≥ [sqrt{ab}]^2

(a + b)^2/4 ≥ ab

(a^2 + 2ab + b^2)/4 ≥ ab

(a^2 + 2ab + b^2) ≥ 4ab

I am stuck here.
 
RTCNTC said:
[(a + b)/2]^2 ≥ [sqrt{ab}]^2

(a + b)^2/4 ≥ ab

(a^2 + 2ab + b^2)/4 ≥ ab

(a^2 + 2ab + b^2) ≥ 4ab

I am stuck here.

What do you get if you subtract 4ab from both sides, and then factor the LHS?
 
(a^2 + 2ab + b^2) ≥ 4ab

a^2 + 2ab + b^2 - 4ab ≥ 0

Factor by grouping.

a^2 + 2ab factors out to be a(a + 2b).

b^2 - 4ab factors out to be b(b - 4a).

a(a + 2b) + b(b - 4a) ≥ 0

(a + b)(a + 2b)(b - 4a) ≥ 0

Do I now set each factor to be ≥ 0?
 
RTCNTC said:
(a^2 + 2ab + b^2) ≥ 4ab

a^2 + 2ab + b^2 - 4ab ≥ 0

What we want to do here is combine like terms:

$$a^2-2ab+b^2\ge0$$

At this point, we should recognize this is the square of a binomial:

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

The square of a real number is always non-negative, so we know that for any valid values of a and b this must be true.
 
  • #10
Thank you for completing this problem for me. Interesting question. From now on, I will post 2 daily questions.
 

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