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

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This answer $\ds{\underline{is\ not}}$ based in 'Induction' as the OP requested.  @RobertZ just called this point to my attention. Indeed, it's a straightforward point of view.

\begin{align}
\color{#f00}{\pars{\sum_{i = 1}^{n}x_{i}}\pars{\sum_{i = 1}^{n}{1 \over xi}}} & =
\sum_{i = 1}^{n}\sum_{k = 1}^{n}{x_{i} \over x_{k}} =
\half\sum_{i = 1}^{n}\sum_{k = 1}^{n}
\pars{{x_{i} \over x_{k}} + {x_{k} \over x_{i}}}
\\[5mm] & =
\half\sum_{i = 1}^{n}\sum_{k = 1}^{n}
\bracks{\pars{\root{x_{i} \over x_{k}} - \root{x_{k} \over x_{i}}}^{2} + 2} \geq
\half\sum_{i = 1}^{n}\sum_{k = 1}^{n}2 = \color{#f00}{n^{2}}
\end{align}
