If $a$ and $b$ are two unequal positive numbers, then: 
$\text{(A) } \frac{2ab}{a+b}>\sqrt{ab}>\frac{a+b}{2}\qquad \text{(B) } \sqrt{ab}>\frac{2ab}{a+b}>\frac{a+b}{2} \\ \text{(C) } \frac{2ab}{a+b}>\frac{a+b}{2}>\sqrt{ab}\qquad \text{(D) } \frac{a+b}{2}>\frac{2ab}{a+b}>\sqrt{ab} \\ \text{(E) } \frac {a + b}{2} > \sqrt {ab} > \frac {2ab}{a + b}$
Using the RMS-AM-GM-HM inequality, we can see that the answer is $\fbox{E}$.