$\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_i}\right) \ge n^2$, for all integers $n\ge 1$. Let $x_1,\ldots, x_n$ be positive integers. Use mathematical induction to prove that
$$\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_i}\right) \ge n^2$$ for all integers $n \ge 1$.
(Given Hint: For all positive integers $a$ and $b$, $\frac{a}{b}+\frac{b}{a} \ge 2$.)
Can anyone help? Thank you.

Hint for the inductive step.
Let $A_n=\sum_{i=1}^{n} x_i$ and let $B_n=\sum_{i=1}^{n} \frac{1}{x_i}$. Then
$$A_{n+1}\cdot B_{n+1}=\left (A_n+ x_{n+1}\right)\cdot \left (B_n+\frac{1}{x_{n+1}} \right)=A_nB_n+\frac{A_n}{x_{n+1}}+B_n x_{n+1}+1\\\geq n^2+1 +\frac{A_n}{x_{n+1}}+B_nx_{n+1}\stackrel{?} {\geq}(n+1)^2=n^2+2n+1.$$
So it suffices to show that
$$\frac{A_n}{x_{n+1}}+B_n x_{n+1}\geq 2n.$$
Now, since $(u+v)\geq 2\sqrt{uv}$, for $u,v>0$,
$$\frac{A_n}{x_{n+1}}+B_n x_{n+1}\geq 2\sqrt{A_n B_n}\geq 2n$$
and we are done.
