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Difference between AM and GM

Source: 1964 AHSME Problem 37

March 19, 2014
algebradifference of squaresspecial factorizationsAMC

Problem Statement

Given two positive number aa, bb such that a<ba<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:
<spanclass=′latex−bold′>(A)</span>(b+a)2ab<spanclass=′latex−bold′>(B)</span>(b+a)28b<spanclass=′latex−bold′>(C)</span>(b−a)2ab<span class='latex-bold'>(A) </span>\dfrac{(b+a)^2}{ab}\qquad<span class='latex-bold'>(B) </span>\dfrac{(b+a)^2}{8b}\qquad<span class='latex-bold'>(C) </span>\dfrac{(b-a)^2}{ab}
<spanclass=′latex−bold′>(D)</span>(b−a)28a<spanclass=′latex−bold′>(E)</span>(b−a)28b<span class='latex-bold'>(D) </span>\dfrac{(b-a)^2}{8a}\qquad <span class='latex-bold'>(E) </span>\dfrac{(b-a)^2}{8b}
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