Difference between revisions of "1964 AHSME Problems/Problem 37"
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==Problem== | ==Problem== | ||
| − | Given two positive | + | Given two positive numbers <math>a</math>, <math>b</math> such that <math>a<b</math>. Let <math>A.M.</math> be their arithmetic mean and let <math>G.M.</math> be their positive geometric mean. Then <math>A.M.</math> minus <math>G.M.</math> 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> | <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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==Solution== | ==Solution== | ||
| − | + | <math>\boxed{ \textbf{(D)} }</math> | |
==See Also== | ==See Also== | ||
Latest revision as of 18:47, 26 January 2021
Problem
Given two positive numbers 
, 
such that 
. Let 
be their arithmetic mean and let 
be their positive geometric mean. Then minus is always less than:
Solution
See Also
| 1964 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 36 |
Followed by Problem 38 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
