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?
 A: Expand it out as $\sum_{i,j} x_i/x_j$, and use $t + 1/t \ge 2$ for $t > 0$
A: Assume known that $(x+y)^2\ge4xy,\forall x,y\in\Bbb R.$
Then suppose this inequality holds for $n.$
Now we have $$\begin{align}(\sum\limits_{i=1}^{n+1}x_i)\cdot(\sum\limits_{i=1}^{n+1}\frac{1}{x_i})&\ge(\sum\limits_{i=1}^{n})\cdot(\sum\limits_{i=1}^{n})+(\sum\limits_{i=1}^{n}x_i)\frac{1}{x_{n+1}}+x_{n+1}(\sum\limits_{i=1}^{n}\frac{1}{x_i})+1\\
&\ge n^2+2\sqrt{((\sum\limits_{i=1}^{n}x_i)\frac{1}{x_{n+1}})\cdot(x_{n+1}(\sum\limits_{i=1}^{n}\frac{1}{x_i}))}+1\\
&\ge n^2+2\sqrt{n^2}+1\\
&=(n+1)^2\end{align}$$
Hope this helps.
A: Let 
$A = \displaystyle \sum_{i=1}^n x_i, B = \displaystyle \sum_{i=1}^n \dfrac{1}{x_i}\implies P_{n+1} = \left(A+x_{n+1}\right)\left(B+\dfrac{1}{x_{n+1}}\right)= P_n+\dfrac{A}{x_{n+1}}+Bx_{n+1}+1\geq n^2+2\sqrt{AB}+1\geq n^2+2n+1 = (n+1)^2$. 
A: Using AM-GM
$$\sum_{k=1}^n x_k\ge n\sqrt[n]{x_1\cdot...\cdot x_n}$$
and
$$\sum_{k=1}^n \dfrac{1}{x_k} \geq n\sqrt[n]{\frac1{x_1}\cdot...\cdot \frac1{x_n}}$$
Then
$$\displaystyle \sum_{k=1}^n x_k\cdot \displaystyle \sum_{k=1}^n \dfrac{1}{x_k} \geq n^2$$
A: 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$


*

*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 $z=1$.

A: By C-S
$$\sum_{i=1}^ns_i\sum_{i=1}^n\frac{1}{s_i}\geq\left(\sum_{i=1}^n\sqrt{s_i}\sqrt{\frac{1}{s_i}}\right)^2=n^2.$$
The Cauchy-Schwarz inequality it's the following.
For all reals $a_i$ and $b_i$ the following inequality holds.
$$\sum_{i=1}^na_i^2\sum_{i=1}^nb_n^2\geq\left(\sum_{i=1}^na_ib_i\right)^2.$$
In our case $a_i=\sqrt{s_i}$ and $b_i=\frac{1}{\sqrt{s_i}}.$
Good luck! 
