Arithmetic and geometric mean I need to prove that for $a=\frac{x+y}{2}$ and $g=\sqrt{xy}$, following statments are true or false:
For $x\not =y,a>g$ and $x=y, a=g$.
I have no idea how to do this, so any help is welcomed. Thanks.
 A: 
$a\ge g \leftrightarrow (x+y-2\sqrt{xy})\ge0 \leftrightarrow (\sqrt{x}-\sqrt{y})^2\ge0$ 

A: Here is a simple proof:
(your question is complete, I don't know why is everyone giving you a hard time)
First consider the case $x = y$ then $ a = \frac {2x} 2 = x = \sqrt {x^2} = g$.
Next consider the case $x \ne y$ then suppose by way of contradiction that $a=g$
then $$ \frac {x+y} {2} = \sqrt {xy}  \iff (x+y)^2 = 4xy  \iff x^2 - 2xy +y^2 =0  \iff (x-y)^2=0 \iff x=y$$
As you can see this leads to a contradiction $ x \ne y \rightarrow a=g \rightarrow x=y $, hence
$$x \ne y \rightarrow a \ne g $$
and our proof is complete.
A: First, we consider 
$$(x-y)^2$$ 
and note that 
$$(x-y)^2 \ge 0$$ for all $x,y \in \mathbb{R}$.
Then 
$$(x-y)^2 = x^2 + y^2 -2xy \ge 0$$
add $4xy$ to both sides of the inequality to get
$$(x+y)^2 \ge 4xy$$
Note that we must not have the case where $xy \not <0$, since we do not have a square root of a negative number.
Ignoring the case above and taking the square root on both sides of the equation we yield the (two) results. (Equality is achieved when $x=y$)
A: Your question is incomplete, Complete question is:
For $x,y \ge 0 $ we have $\frac{x+y}{2} \ge \sqrt{xy}$
For proving this we have to considering:
$$(x-y)^2$$
then
$$(x-y)^2 \ge 0 $$
because it's square! and we know $$(x-y)^2 = x^2-2xy+y^2$$
then
$$x^2-2xy+y^2 \ge 0$$
now plus $4xy$ to both sides:
$$x^2+2xy+y^2 \ge 4xy$$
then
$$(x+y)^2 \ge 4xy$$
then
$$|(x+y)| \ge 2\sqrt{xy}$$
because $x,y \ge 0$ then $|(x+y)| = (x+y)$
and now
$$\frac{x+y}{2} \ge \sqrt{xy}$$
