Prove that $\left(\sum^n_{k=1}x_k\right)\left(\sum^n_{k=1}y_k\right)\geq n^2$ 
If $x_1,...,x_n$ are positive real numbers and if $y_k=1/x_k$, prove that $$\left(\sum^n_{k=1}x_k\right)\left(\sum^n_{k=1}y_k\right)\geq n^2.$$

I've been learning induction, and I've come across this problem that I really can't even break down and begin to think about. I've been told it has something to do with Cauchy-Schwarz, but I cannot figure out how to apply it. I would appreciate help figuring out how to go about and formulate this proof. Thanks!
 A: There are couple of ways to prove this. One way is via AM-GM, i.e., we have
$$\sum_{k=1}^n x_k \geq n \sqrt[n]{x_1 x_2 \ldots x_n}$$
and
$$\sum_{k=1}^n \dfrac1{x_k} \geq n \sqrt[n]{\dfrac1{x_1 x_2 \ldots x_n}}$$
Multiplying the two, we get what we want.
Another way is consider the vectors
$$\left(\sqrt{x_1},\sqrt{x_2}, \ldots, \sqrt{x_n} \right) \text{ and }\left(\dfrac1{\sqrt{x_1}},\dfrac1{\sqrt{x_2}}, \ldots, \dfrac1{\sqrt{x_n}} \right)$$ and apply Cauchy-Schwarz to get what you want.
A: You can in fact prove it by induction on $n$. For the induction step observe that
$$\begin{align*}
\left(\sum_{k=1}^{n+1}x_k\right)\left(\sum_{k=1}^{n+1}\frac1{x_k}\right)&=\left(\sum_{k=1}^nx_k+x_{n+1}\right)\left(\sum_{k=1}^n\frac1{x_k}+\frac1{x_{n+1}}\right)\\
&=\left(\sum_{k=1}^nx_k\right)\left(\sum_{k=1}^n\frac1{x_k}\right)+x_{n+1}\sum_{k=1}^n\frac1{x_k}+\frac1{x_{n+1}}\sum_{k=1}^nx_k+1\\
&\ge n^2+1+x_{n+1}\sum_{k=1}^n\frac1{x_k}+\frac1{x_{n+1}}\sum_{k=1}^nx_k\;;
\end{align*}$$
$(n+1)^2=n^2+2n+1$, so to finish the step, it suffices to show that
$$x_{n+1}\sum_{k=1}^n\frac1{x_k}+\frac1{x_{n+1}}\sum_{k=1}^nx_k\ge 2n\;.$$
For $k=1,\ldots,n$ let $u_k=\dfrac{x_k}{x_{n+1}}$; then
$$x_{n+1}\sum_{k=1}^n\frac1{x_k}+\frac1{x_{n+1}}\sum_{k=1}^nx_k=\sum_{k=1}^n\left(\frac1{u_k}+u_k\right)\;,$$
and it’s not hard to show that if $u>0$, then $\dfrac1u+u\ge 2$, either by showing that $f(u)=\frac1u+u$ on the positive reals has a minimum at $u=1$, or by observing that for $u>0$ we have $f(u)\ge 2$ if and only if $u^2+1\ge 2u$ and showing that this inequality is always true.
A: The function 
$$f(x)=\frac{1}{x}$$
Is convex on $(0,\infty)$, so for $a_1,\ldots,a_n>0$ we have
$$\frac{n}{a_1+\cdots+a_n}\leq\frac{1}{na_1}+\cdots+\frac{1}{na_n}.$$
Rearranging, we obtain the result.
A: $$\left(\sum_{k=1}^n {\sqrt{x_k}}^2\right)\left(\sum_{k=1}^n {\sqrt{x_k^{-1}}}^2\right)
\ge \left(\sum_{k=1}^n {\sqrt{x_k}}\sqrt{x_k^{-1}}\right)^2 = n^2$$
A: We let $x_{i+n}=x_i$ for $i\in\{1,2\dots n\}$ .What you have is $\sum_\limits{i=1}^n (\frac{x_1}{x_{1+i}}+\frac{x_2}{x_{2+i}}+\dots \frac{x_n}{x_{n+i}}).$
By the rearrangement inequality each of these summands is greater than $1+1+\dots +1=n$
A: \begin{align}
\sum_{k = 1}^{n}\left(\sqrt{x_{k}\,} + {\mu \over \sqrt{x_{k}\,}}\right)^{2} &\geq 0\,,
\qquad
\mu \in {\mathbb R}
\\[3mm]
\left(\sum_{k = 1}^{n}{1 \over x_{k}}\right)\mu^{2}
+
2n\mu
+
\left(\sum_{k = 1}^{n}x_{k}\right) &\geq 0
\\
\left(2n\right)^{2}
-
4\left(\sum_{k = 1}^{n}{1 \over x_{k}}\right)\left(\sum_{k = 1}^{n}x_{k}\right)
&
\leq 0
\end{align}
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
\sum_{k = 1}^{n}x_{k}\sum_{k = 1}^{n}{1 \over x_{k}}
\color{#000000}{\ \geq\ }
n^{2}
\quad}
\\ \\ \hline
\end{array}
$$
A: You can use Cauchy-Schwarz inequality with $u_i = \sqrt{x_i}$ and $v_i=\frac{1}{\sqrt{x_i}}$.
A: By Chebyshev's Inequality we have that
$$\left(\sum_{k=1}^n x_k\right)\cdot\left(\sum_{k=1}^n \frac{1}{x_k}\right) \geq n\sum_{k=1}^n x_k \frac{1}{x_k}=n^2$$
and therefore
$$\left(\frac{1}{n}\sum_{k=1}^n x_k\right)\cdot\left(\frac{1}{n}\sum_{k=1}^n \frac{1}{x_k}\right) \geq 1$$
