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

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The discussion explores the relationship between the geometric mean (G.M.) and arithmetic mean (A.M.) for both numerical and algebraic expressions. It establishes that for two positive numbers, A.M. is greater than or equal to G.M., with equality occurring only when the numbers are equal. The participants analyze specific cases, such as a = 1 and b = 2, and algebraic expressions like a = x and b = 1/x, to illustrate these concepts. The proof involves manipulating inequalities and recognizing that the square of a binomial is always non-negative. The conclusion reinforces the validity of A.M. being greater than G.M. for positive values.
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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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