What is this inequality called: $x + y \geq 2\sqrt{xy}$? 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?
 A: 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.
A: 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 $p=q=2$.
A: 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 $x_1=\cdots=x_n$.
A: 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 $\,x\,,\,y\geq 0\,$ .
