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!

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.

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.