Difference between revisions of "Solution to AM - GM Introductory Problem 2"
(Created page with "===Problem=== Find the maximum of <math>2 - a - \frac{1}{2a}</math> for all positive <math>a</math>. ===Solution=== We can rewrite the given expression as <math>2 - (a + \f...") |
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By AM - GM, the arithmetic mean of <math>a</math> and <math>\frac{1}{2a}</math> is at least their geometric mean, or <math>\frac{\sqrt{2}}{2}</math>. This means the sum of <math>a</math> and <math>\frac{1}{2a}</math> is at least <math>\sqrt{2}</math>. We can prove that we can achieve this minimum for <math>a</math> + <math>\frac{1}{2a}</math> by plugging in <math>a = \frac{\sqrt{2}}{2}</math> by solving <math>a + \frac{1}{2a} = \sqrt{2}</math> for <math>a</math>. | By AM - GM, the arithmetic mean of <math>a</math> and <math>\frac{1}{2a}</math> is at least their geometric mean, or <math>\frac{\sqrt{2}}{2}</math>. This means the sum of <math>a</math> and <math>\frac{1}{2a}</math> is at least <math>\sqrt{2}</math>. We can prove that we can achieve this minimum for <math>a</math> + <math>\frac{1}{2a}</math> by plugging in <math>a = \frac{\sqrt{2}}{2}</math> by solving <math>a + \frac{1}{2a} = \sqrt{2}</math> for <math>a</math>. | ||
− | Plugging in <math>a = \frac{\sqrt{2}}{2}</math> | + | Plugging in <math>a = \frac{\sqrt{2}}{2}</math> into our original expression that we wished to maximize, we get that <math>2 - a - \frac{1}{2a} = \boxed{2 - \sqrt{2}}</math>, which is our answer. |
Revision as of 21:15, 13 May 2020
Problem
Find the maximum of for all positive .
Solution
We can rewrite the given expression as . To maximize the whole expression, we must minimize . Since is positive, so is . This means AM - GM will hold for and .
By AM - GM, the arithmetic mean of and is at least their geometric mean, or . This means the sum of and is at least . We can prove that we can achieve this minimum for + by plugging in by solving for .
Plugging in into our original expression that we wished to maximize, we get that , which is our answer.