Prove directly that for $x,y \ge 0$ The question that I am stuck on is as follows:
    Prove directly that for $x,y \ge 0$
    $\sqrt{xy}\le (x+y)/2$.
When does equality hold?
I have been working at it for almost 20 minutes. Can anyone help me complete the problem? Thanks a lot!
 A: This is equivalent, via squaring, to showing that
$$xy \le \frac{(x + y)^2}{4}$$
This can be rearranged, and is equivalent to showing that
$$x^2 + 2xy + y^2 \ge 4xy$$
which is equivalent to
$$x^2 - 2xy + y^2 \ge 0$$
Finally, this is simply the statement $$(x - y)^2 \ge 0$$ which is obviously true. We also note that equality holds here if and only if $x = y$.
A: There are lots of ways to do this.  Here's one I figured out before I knew much of anything, when I was about 14:
Suppose $0<x<y$.  What do you have to multiply $x$ by to get $\sqrt{xy\,{}}$?  And what do you have to multiply $\sqrt{xy\,{}}$ by to get $y$?  It's the same thing both times.  Call it $c$.  If multiplying $x$ by $c$ gave you a number at least as big as $(x+y)/2$, then multiplying by $c$ again would give you something bigger than $y$, because the thing you're multiplying by $c$ is bigger this time.
That shows that equality can hold ONLY when $x=y$.
A: It's an inequality. Note that 
$(x-y)^{2}\geq 0$. Therefore $x^{2}-2xy+y^{2}\geq 0\Rightarrow x^{2}+2xy+y^{2}\geq 4xy\Rightarrow (x+y)^{2}\geq 4xy$. Then you are done. 
This is an example of the arithmetic mean is always greater than the geometric mean. 
