1952 AMC 12 Problem 45

Attempt Problem 45 of the 1952 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 1952 AMC 12 solutions, or check the answer key.

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45.

If aa and bb are two unequal positive numbers, then:
2aba+b>ab>a+b2\dfrac{2ab}{a+b}\gt\sqrt{ab}\gt\dfrac{a+b}{2}
ab>2aba+b>a+b2\sqrt{ab}\gt\dfrac{2ab}{a+b}\gt\dfrac{a+b}{2}
2aba+b>a+b2>ab\dfrac{2ab}{a+b}\gt\dfrac{a+b}{2}\gt\sqrt{ab}
a+b2>2aba+b>ab\dfrac{a+b}{2}\gt\dfrac{2ab}{a+b}\gt\sqrt{ab}
a+b2>ab>2aba+b\dfrac{a+b}{2}\gt\sqrt{ab}\gt\dfrac{2ab}{a+b}
Answer: E
Concepts:AM-GM Inequalityharmonic meaninequality
Small hint:
Recognize the arithmetic, geometric, and harmonic means of aa and bb
Big hint:
For unequal positive numbers, the mean inequalities are strict
Solution:
The arithmetic-geometric mean inequality gives a+b2>ab \frac{a+b}{2}\gt\sqrt{ab} for unequal positive a,b.a,b. Applying the same inequality to 1a\frac{1}{a} and 1b\frac{1}{b} gives ab>2aba+b. \sqrt{ab}\gt\frac{2ab}{a+b}. Therefore the decreasing order is arithmetic mean, geometric mean, harmonic mean.
Thus, the correct answer is E.
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