Proving the given inequality Let $a_1,...,a_n$ be positive numbers. Show that $(a_1+...+a_n)(\frac{1}{a_1}+...+\frac{1}{a_n})\ge n^2$. Just not really sure how to approach this problem, any tips/solutions would be greatly appreciated.
 A: The Cauchy-Schwarz Inequality can be stated as follows:
$$(x_1^2+x_2^2+\dots+x_n^2)(y_1^2+y_2^2+\dots+y_n^2) \ge(x_1y_1+x_2y_2+\dots+x_ny_n)^2$$
Now taking $x_i = \sqrt {a_i}$ and $y_i = \frac 1 {\sqrt {a_i}}$ (which is valid since each $a_i > 0$):
$$(\sqrt {a_1}^2+\sqrt {a_2}^2+\dots+\sqrt {a_n}^2)\left(\frac 1 {\sqrt {a_1}^2}+\frac 1 {\sqrt {a_2}^2}+\dots+\frac 1 {\sqrt {a_n}^2}\right)\ge\left(\frac {\sqrt{a_1}}{\sqrt{a_1}}+\dots+\frac {\sqrt{a_n}}{\sqrt{a_n}}\right)^2$$
which yields your inequality after simplification.
A: You should probablly use either the C-S inequality or you can devide the set of all a_i into bigger than one and smaller than one and then neglacte them from the opposite sum (the a's that are smaller than one neglect from the sum of a's to the power of one and vice versa)
A: You are going to see which Inequality suits this problem, once it is written in this form:
$$
\tag{1}\label{1}\frac{a_1 + a_2 + \cdots + a_n}{n} \geqslant \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}}
$$
Equivalently,
$$
\left(a_1 + a_2 + \cdots + a_n\right)
\left(\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}\right) \geqslant n^2
$$
