Prove that if $\displaystyle 0 \leq a<b$, then

$\displaystyle a < \sqrt{ab} < \frac{a+b}{2} < b$

I know how to find $\displaystyle a < \sqrt{ab}$, but I don't understand how to find $\displaystyle \sqrt{ab} < \frac{a+b}{2}$, or that that will be less than b.

Can this even be proved? It doesn't appear to be true.