Please 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