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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>.
  
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Plugging in <math>a = \frac{\sqrt{2}}{2}</math> for 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.
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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 $2 - a - \frac{1}{2a}$ for all positive $a$.

Solution

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

By AM - GM, the arithmetic mean of $a$ and $\frac{1}{2a}$ is at least their geometric mean, or $\frac{\sqrt{2}}{2}$. This means the sum of $a$ and $\frac{1}{2a}$ is at least $\sqrt{2}$. We can prove that we can achieve this minimum for $a$ + $\frac{1}{2a}$ by plugging in $a = \frac{\sqrt{2}}{2}$ by solving $a + \frac{1}{2a} = \sqrt{2}$ for $a$.

Plugging in $a = \frac{\sqrt{2}}{2}$ into our original expression that we wished to maximize, we get that $2 - a - \frac{1}{2a} = \boxed{2 - \sqrt{2}}$, which is our answer.

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