Inequality : $\sum_{k=1}^n x_k\cdot \sum_{k=1}^n \frac{1}{x_k} \geq n^2$ I have to show the inequality of $$\left(\sum_{i=1}^n x_i\right)*\left(\sum_{i=1}^n \frac{1}{x_i}\right) \geq n^2.$$For $x_1, ... x_n \in \mathbb{R_{>0}}$ and $ n \geq 1$. I wanted to show this inequality by induction. The basis is clear, but I am not sure how to do the inductive step. Can someone help me with that?

Using Induction on $n$, $\sum_{i=1}^{n+1}x_i$.$\sum_{i=1}^{n+1}1/x_i$$\geq n^2+x_{n+1}$$(\frac{1}{x_1}+$$\frac{1}{x_2}+....+$$\frac{1}{x_n}$)$+$$\frac{1}{x_{n+1}}$$(x_1+x_2+......+x_n)$
   $=n^2+1+(\frac{x_{n+1}}{x_1}+\frac{x_1}{x_{n+1}})+....+ (\frac{x_{n+1}}{x_n}+\frac{x_n}{x_{n+1}})$$\geq n^2+1+2n=(n+1)^2$


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*Note: The minimum value of each term of type $(\frac{x_{n+1}}{x_i}+\frac{x_i}{x_{n+1}})$; $1\leq i\leq n$ is $2$ as you can see that $f(x,y)=\frac{x}{y}+\frac{y}{x}$ in one variable $z=\frac{x}{y}$ gives $f(z)=z+\frac{1}{z}$; $z>0$. With the help of second derivative test, you can easily verify that $f(z)$ has minimum value $2$ at $z=1$.

