1964 AMC 12 Problem 37

Question

Given two positive numbers

a,

b

such that

a\lt b.

Let A.M. be their arithmetic mean and let G.M. be their positive geometric mean. Then A.M. minus G.M. is always less than:

\dfrac{(b+a)^2}{ab}

\dfrac{(b+a)^2}{8b}

\dfrac{(b-a)^2}{ab}

\dfrac{(b-a)^2}{8a}

\dfrac{(b-a)^2}{8b}

Rewrite

\frac{a+b}{2}-\sqrt{ab}

using

\sqrt b-\sqrt a

Factor

b-a=(\sqrt b-\sqrt a)(\sqrt b+\sqrt a)

to compare with choice D

We have

\frac{a+b}{2}-\sqrt{ab} =\frac{(\sqrt b-\sqrt a)^2}{2}.

After canceling the positive factor

(\sqrt b-\sqrt a)^2,

comparison with choice D reduces to

\frac12 <\frac{(\sqrt b+\sqrt a)^2}{8a}.

This is true because

\sqrt b+\sqrt a\gt2\sqrt a.

Thus A.M. minus G.M. is always less than the expression in choice D.

Therefore, the correct answer is D.

Problem 37 in other years