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Apparently it's a famous inequality taught in 1st year calculus but I have never even seen it before nor know it has a name.

$x + y \geq 2\sqrt{xy}$

It looks like it is just saying $(x + y)^2 \geq 4xy$, so it's somehow derived from sum of squares?

What is the name (if it even has one) of this inequality?

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i never heard about this fame... –  Integral Sep 15 '12 at 2:35
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5 Answers

up vote 7 down vote accepted

This is called the AM-GM or Arithmetic Mean-Geometric Mean inequality. It generalizes to $$\frac{x_1+\cdots+x_n}{n}\geq \sqrt[n]{x_1\cdots x_n}$$ and more information on it can be found on Wikipedia.

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AHHHH That's the name! –  sidht Sep 15 '12 at 3:17
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In addition to being an example of the AM-GM inequality as explained in other answers, it can also be interpreted as a special case of Young's inequality: $$ab \le \frac{a^p}{p}+\frac{b^q}{q} \text{, where } \frac{1}{p}+\frac{1}{q} = 1$$

Set $a=\sqrt x$, $b=\sqrt y$, and $p=q=2$.

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For positive numbers, the arithmetic mean is at least as big as the geometric mean. The arithmetic mean is $$ \frac{x_1+\cdots+x_n}{n} $$ and the geometric mean is $$ (x_1\cdots x_n)^{1/n}. $$ The inequality you've written is the special case in which $n=2$.

See this article: http://en.wikipedia.org/wiki/Arithmetic-geometric_mean_inequality

Later note: The two means are equal only if $x_1=\cdots=x_n$.

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Well, after one gets into this inequality soul it can even be called "trivial", as you can check that

$$x+y\geq 2\sqrt{xy}\Longleftrightarrow (x+y)^2\geq 4xy\Longleftrightarrow (x-y)^2\geq 0$$

Of course, the basic assumption is $\,x\,,\,y\geq 0\,$ .

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