TITLE: Proving the AM-GM inequality for 2 numbers $\sqrt{xy}\le\frac{x+y}2$
QUESTION [11 upvotes]: I am having trouble with this problem from my latest homework.
Prove the arithmetic-geometric mean inequality. That is, for two positive real
numbers $x,y$, we have
$$ \sqrt{xy}≤ \frac{x+y}{2} .$$
Furthermore, equality occurs if and only if $x = y$.
Any and all help would be appreciated.

REPLY [6 votes]: I am surprised no one has given the following very straightforward algebraic argument:
\begin{align}
0\leq(x-y)^2&\Longleftrightarrow 0\leq x^2-2xy+y^2\tag{expand}\\[0.5em]
&\Longleftrightarrow 4xy\leq x^2+2xy+y^2\tag{add $4xy$ to both sides}\\[0.5em]
&\Longleftrightarrow xy\leq\left(\frac{x+y}{2}\right)^2\tag{div. sides by 4 & factor}\\[0.5em]
&\Longleftrightarrow \sqrt{xy}\leq\frac{x+y}{2}.\tag{since $x,y\in\mathbb{R}^+$}
\end{align}
In regards to equality, notice that $\sqrt{xy}\leq\frac{x+y}{2}\leftrightarrow 2\sqrt{xy}\leq x+y$, and it becomes clear that equality holds if and only if $x=y$. $\blacksquare$