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

[mathdad]

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. ?

 

[MarkFL]

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:

\(\displaystyle \sqrt{2}<\frac{3}{2}\)

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

 

[mathdad]

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?

 

[MarkFL]

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

 

[mathdad]

Are you saying to treat it like a radical equation?

 

[mathdad]

[(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.

 

[MarkFL]

[(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?

 

[mathdad]

(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?

 

[MarkFL]

(a^2 + 2ab + b^2) ≥ 4ab

a^2 + 2ab + b^2 - 4ab ≥ 0

What we want to do here is combine like terms:

\(\displaystyle a^2-2ab+b^2\ge0\)

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

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

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.

 

[mathdad]

Thank you for completing this problem for me. Interesting question. From now on, I will post 2 daily questions.