Difference between revisions of "1964 AHSME Problems/Problem 37"

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==Problem==
 
==Problem==
  
Given two positive number <math>a</math>, <math>b</math> such that <math>a<b</math>. Let A.M. be their arithmetic mean and let <math>G.M. be their positive geometric mean. Then A.M. minus G.M. is always less than:
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Given two positive number <math>a</math>, <math>b</math> such that <math>a<b</math>. 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:
  
</math>\textbf{(A) }\dfrac{(b+a)^2}{ab}\qquad\textbf{(B) }\dfrac{(b+a)^2}{8b}\qquad\textbf{(C) }\dfrac{(b-a)^2}{ab}<math>
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<math>\textbf{(A) }\dfrac{(b+a)^2}{ab}\qquad\textbf{(B) }\dfrac{(b+a)^2}{8b}\qquad\textbf{(C) }\dfrac{(b-a)^2}{ab}</math>
  
</math>\textbf{(D) }\dfrac{(b-a)^2}{8a}\qquad \textbf{(E) }\dfrac{(b-a)^2}{8b}$
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<math>\textbf{(D) }\dfrac{(b-a)^2}{8a}\qquad \textbf{(E) }\dfrac{(b-a)^2}{8b}</math>

Revision as of 17:21, 5 March 2014

Problem

Given two positive number $a$, $b$ such that $a<b$. 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:

$\textbf{(A) }\dfrac{(b+a)^2}{ab}\qquad\textbf{(B) }\dfrac{(b+a)^2}{8b}\qquad\textbf{(C) }\dfrac{(b-a)^2}{ab}$

$\textbf{(D) }\dfrac{(b-a)^2}{8a}\qquad \textbf{(E) }\dfrac{(b-a)^2}{8b}$

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