Inequality $( \sum_{i=1}^{n} a_i)( \sum_{i=1}^{n} \frac{1}{a_i})\geq n^2$

Question

For $a_1, ... , a_2 > 0$, I need to prove that $( \sum_{i=1}^{n} a_i)( \sum_{i=1}^{n} \frac{1}{a_i})\geq n^2 $.

When do we have inequality?

Should I do it by induction? any hints

Answer 1

Hint: Take $y_{i} = \sqrt{a_{i}}$ and $x_{i} = \frac{1}{\sqrt{a_{i}}}$ in the Cauchy-Schwarz inequality.

Then $n^{2} = \left(\sum_{i=1}^{n}x_{i}y_{i} \right)^{2} \leq \left (\sum_{i=1}^{n}x_{i}^{2} \right) \cdot \left(\sum_{i=1}^{n} y_{i}^{2} \right)$

Answer 2

Apply AM-GM inequality twice: $$LHS \geq n\sqrt[n]{a_1a_2\cdots a_n}\cdot n\sqrt[n]{\dfrac{1}{a_1a_2\cdots a_n}}=n^2=RHS$$, and equality occurs when $a_1=a_2=\cdots = a_n$.

Answer 3

Equivalently, you have to prove that $$ \sum_{1\le i ,j\le n: i\neq j}\frac{a_i}{a_j} \ge n^2-n. $$ Now, $x+\frac{1}{x}\ge 2$ by am-gm, therefore $$ \sum_{1\le i <j\le n}\frac{a_i}{a_j}+\frac{a_j}{a_i}\ge 2\binom{n}{2}=n^2-n. $$