$\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_i}\right) \ge n^2$, 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.
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}
A: 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.
A: 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 $n+1$.
