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Can anyone explain me what is AM-GM inequality?

Can anyone explain me what is AM-GM inequality?

Grade:10

1 Answers

Chetan Mandayam Nayakar
312 Points
12 years ago

The most basic arithmetic mean-geometric mean (AM-GM) inequality states simply that if $ x$ and $ y$ are nonnegative real numbers, then $ (x+y)/2 \ge \sqrt{xy}$, with equality if and only if $ x=y$. The last phrase ``with equality...' means two things: first, if $ x=y \ge 0$, then $ (x+y)/2 = \sqrt{xy}$ (obvious); and second, if $ (x+y)/2 = \sqrt{xy}$ for some $ x,y \ge 0$, then $ x=y$. It follows that if $ x,y \ge 0$ and $ x \not= y$, then inequality is strict: $ (x+y)/2 >\sqrt{xy}$.

Here is a one-line proof of the AM-GM inequality for two variables:

 

 

$\displaystyle \frac{x+y}{2}-\sqrt{xy} = \frac{1}{2} \left(\sqrt{x}-\sqrt{y}\right)^2 \ge 0.$

 

 

The AM-GM inequality generalizes to $ n$ nonnegative numbers.


AM-GM inequality:
If $ x_1,\dots,x_n \ge 0$, then

 

 

$\displaystyle \frac{x_1+x_2+\dots+x_n}{n} \ge \sqrt[n]{x_1 x_2 \dots x_n}$

 

 

with equality if and only if

 

$ x_1=x_2=\dots=x_n$.

 

 

 

 

 

 

 

 

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