How to prove $(a_1+a_2+\dots a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}\right)\ge n^2$? let $a_1,a_2,\dots ,a_n$ be a positive real numbers . prove that 
$$(a_1+a_2+\dots a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}\right)\ge n^2$$
how to prove this inequality ..is this result prove by Induction 

Another proof: See Art of Problem Solving about Chebyshev's inequality.
There are two different versions of it. Use the second version.
If $a_1\ge a_2\ge \ldots \ge a_n$ and $b_n\ge b_{n-1}\ge \ldots \ge b_1$, then $$n\left(\sum_{i=1}^{n} a_ib_i\right)\le \left(\sum_{i=1}^{n}a_i\right)\left(\sum_{i=1}^{n}b_i\right)$$
In this case, you're given $a_1,a_2,\ldots,a_n>0$. Use the inequality.
Notice that if WLOG (without loss of generality) $a_1\ge a_2\ge \ldots\ge a_n$, then $\frac{1}{a_n}\ge \frac{1}{a_{n-1}}\ge \ldots \ge \frac{1}{a_1}$.
