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1964 AMC 12 Problem 37

Attempt Problem 37 of the 1964 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1964 AMC 12 solutions, or check the answer key.

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37.

Given two positive numbers a,a, bb such that a<b.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:

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

(b+a)28b\dfrac{(b+a)^2}{8b}

(b−a)2ab\dfrac{(b-a)^2}{ab}

(b−a)28a\dfrac{(b-a)^2}{8a}

(b−a)28b\dfrac{(b-a)^2}{8b}

Answer: D
Concepts:AM-GM Inequalityinequalityradical
Difficulty rating: 1670
Small Hint:

Rewrite a+b2−ab\frac{a+b}{2}-\sqrt{ab} using b−a\sqrt b-\sqrt a

Big Hint:

Factor b−a=(b−a)(b+a)b-a=(\sqrt b-\sqrt a)(\sqrt b+\sqrt a) to compare with choice D

Solution:

We have a+b2−ab=(b−a)22. \frac{a+b}{2}-\sqrt{ab} =\frac{(\sqrt b-\sqrt a)^2}{2}. After canceling the positive factor (b−a)2,(\sqrt b-\sqrt a)^2, comparison with choice D reduces to 12<(b+a)28a. \frac12 <\frac{(\sqrt b+\sqrt a)^2}{8a}. This is true because b+a>2a.\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.

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