Prove the inequality $\sum_{1\le iPlease demonstrate this is true
This is the exercise: 
$$\sqrt{a_1a_2}+\sqrt{a_1a_3}+\ldots+\sqrt{a_1a_n}+\sqrt{a_2a_3}+\ldots+\sqrt{a_{n-1}a_n}<\frac{n-1}2(a_1+a_2+a_3+\ldots+a_n).$$
I tried to solve it, but I couldn't do anything right.
This is my idea:
$\sqrt{a_1a_2}<\frac{a_1+a_2}2$ -because geometric mean < arithmetic mean;
$\sqrt{a_1a_3}<\frac{a_1+a_3}2$,
$\dots$,
$\sqrt{a_{n-1}a_n}<\frac{a_{n-1}+a_n}2$
Please give me the answer!!
$a_1,a_2,a_3,\ldots,a_n$ are real, positive nummbers

Or
$$
\begin{eqnarray}
0 & \leq &\textrm{variance of the sequence } \sqrt{a_1}, \sqrt{a_2}, \dotsc, \sqrt{a_n} \\
 & = &\frac{\sum_k a_k}{n} - \left( \frac{\sum_k \sqrt{a_k}}{n} \right)^2 \\
 & = & \frac{n-1}{n^2} \sum_k a_k - \frac{2}{n^2} \sum_{j < k}\sqrt{a_ja_k}
\end{eqnarray}
$$
