$\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_i}\right) \ge n^2$, 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.

Suppose the given inequality holds for an integer $n(\geq 1)$.
Then for $n+1$,
$$\left(\sum_{i=1}^{n+1} x_i\right)\left(\sum_{i=1}^{n+1} \frac{1}{x_i}\right)=\left(\sum_{i=1}^{n} x_i+x_{n+1}\right)\left(\sum_{i=1}^{n} \frac{1}{x_i}+\frac{1}{x_{n+1}}\right)$$
Expanding this gives us
$$=\left(\sum_{i=1}^{n} x_i\right)\left(\sum_{i=1}^{n} \frac{1}{x_i}\right)+x_{n+1}\sum_{i=1}^{n} \frac{1}{x_i}+\frac{1}{x_{n+1}}\sum_{i=1}^{n} x_i+1$$
Since the given inequality holds for the integer $n$, the first term is larger than $n^2$. So,
$$\geq n^2+x_{n+1}\sum_{i=1}^{n} \frac{1}{x_i}+\frac{1}{x_{n+1}}\sum_{i=1}^{n} x_i+1$$
Now we show that $x_{n+1}\sum_{i=1}^{n} \frac{1}{x_i}+\frac{1}{x_{n+1}}\sum_{i=1}^{n} x_i$ is larger than $2n$.
This is where we use the hint. Expanding the $\sum$ gives us
$$x_{n+1}\left(\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}\right)+\frac{1}{x_{n+1}}\left(x_1+x_2+\cdots+x_n\right)$$
Rearranging the terms, we get
$$\left( \frac{x_{n+1}}{x_1}+\frac{x_1}{x_{n+1}} \right)+\left( \frac{x_{n+1}}{x_2}+\frac{x_2}{x_{n+1}} \right)+\cdots+\left( \frac{x_{n+1}}{x_n}+\frac{x_n}{x_{n+1}} \right)$$
and using the hint, this is larger than $2n$
Thus,
$$\left(\sum_{i=1}^{n+1} x_i\right)\left(\sum_{i=1}^{n+1} \frac{1}{x_i}\right)\geq n^2+2n+1=(n+1)^2$$
The inequality holds for $n+1$.
