Difference between revisions of "Solution to AM - GM Introductory Problem 1"
Angrybird029 (talk | contribs) (Created page with "==Problem== For nonnegative real numbers <math>a_1,a_2,\cdots a_n</math>, demonstrate that if <math>a_1a_2\cdots a_n=1</math> then <math>a_1+a_2+\cdots +a_n\ge n</math>. ==Sol...") |
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For nonnegative real numbers <math>a_1,a_2,\cdots a_n</math>, demonstrate that if <math>a_1a_2\cdots a_n=1</math> then <math>a_1+a_2+\cdots +a_n\ge n</math>. | For nonnegative real numbers <math>a_1,a_2,\cdots a_n</math>, demonstrate that if <math>a_1a_2\cdots a_n=1</math> then <math>a_1+a_2+\cdots +a_n\ge n</math>. | ||
==Solution== | ==Solution== | ||
| − | + | Since <math>a_1a_2\cdots a_n=1</math>, the geometric mean (<math>\sqrt[n]{a_1a_2\cdots a_n}</math>) must also equal <math>1</math>. | |
| + | |||
| + | The AM-GM Inequality states that the arithmetic mean of a set of non-negative numbers is greater than or equal to the geometric mean, so that means that <math>\frac{a_1+a_2+\cdots +a_n}{n}\geq 1</math>. | ||
| + | |||
| + | Rearranging, we get <math>a_1+a_2+\cdots +a_n\ge n</math>, as required. <math>\square</math> | ||
Back to the [[Arithmetic Mean-Geometric Mean Inequality]]. | Back to the [[Arithmetic Mean-Geometric Mean Inequality]]. | ||
Latest revision as of 12:10, 6 March 2021
Problem
For nonnegative real numbers 
, demonstrate that if 
then 
.
Solution
Since , the geometric mean (![$\sqrt[n]{a_1a_2\cdots a_n}$](http://latex.artofproblemsolving.com/5/5/6/556f366fbe0e31be453f8b9c4ea6da4572cc66b4.png)
) must also equal 
.
The AM-GM Inequality states that the arithmetic mean of a set of non-negative numbers is greater than or equal to the geometric mean, so that means that 
.
Rearranging, we get , as required. 
Back to the Arithmetic Mean-Geometric Mean Inequality.
