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

As $a_i$ s  $>0$
$$(\sqrt {a_i}-\sqrt {a_j})^2\ge0\implies a_i+a_j\ge 2\sqrt{a_ia_j}$$ 
Putting $i=1,2,\cdots,n$ and $j=1,2,\cdots, n$  and adding them we get,
$$(n+1)\sum a_i\ge \sum2\sqrt{a_ia_j}+\sum 2a_i(\text{ this is due to } i=j)$$
$$\implies (n-1)\sum a_i\ge 2\sum\sqrt{a_ia_j}$$
