# Difference between revisions of "Solution to AM - GM Introductory Problem 2"

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We can rewrite the given expression as <math>2 - (a + \frac{1}{2a})</math>. To maximize the whole expression, we must minimize <math>a + \frac{1}{2a}</math>. Since <math>a</math> is positive, so is <math>\frac{1}{2a}</math>. This means AM - GM will hold for <math>a</math> and <math>\frac{1}{2a}</math>. | We can rewrite the given expression as <math>2 - (a + \frac{1}{2a})</math>. To maximize the whole expression, we must minimize <math>a + \frac{1}{2a}</math>. Since <math>a</math> is positive, so is <math>\frac{1}{2a}</math>. This means AM - GM will hold for <math>a</math> and <math>\frac{1}{2a}</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 | + | 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 + \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> 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. | 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. |

## Latest revision as of 21:18, 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.