How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$? 
How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?

im trying to use AM-GM 
$\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}}
  \le \sum_{k=1}^n \frac{{\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}}{n}$
$ln\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}} \le ln \frac{1}{n} \sum_{k=1}^n {\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}$
$ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}\le  \sum_{k=1}^n {\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}$
im not sure is this right or not, however i dont know how to include the $n^2$ in the nominator?
is there also alternative proof using jensen inequality?
 A: The inequality you are interested in is equivalent $$\frac{1}{\frac{x_1+x_2+\dots+x_n}{n}} \le \frac{\frac{1}{x_1}+ \frac{1}{x_2} + \dots+ \frac{1}{x_n}}{n}.$$which is just AM-HM inequality.
You can also prove it using CS inequality.
A: For $x_i>0$ with $i=1,\dots,n$, the given inequality is equivalent to
$$\frac{1}{\frac{x_1+x_2+\dots+x_n}{n}} \le \frac{\frac{1}{x_1}+ \frac{1}{x_2} + \dots+ \frac{1}{x_n}}{n}.$$
Now note that $x\to 1/x$ is convex in $(0,+\infty)$ and use Jensen's inequality.
A: It's wrong. Try $x_1\rightarrow0^-$
For positive variables it's true by C-S:
$$\sum_{k=1}^nx_k\sum_{k=1}\frac{1}{x_k}\geq\left(\sum_{k=1}^n\sqrt{x_k\cdot\frac{1}{x_k}}\right)^2=n^2.$$
Also, AM-GM works:
$$\sum_{k=1}^nx_k\sum_{k=1}\frac{1}{x_k}\geq n\sqrt[n]{\prod_{k=1}^nx_k}\cdot n\sqrt[n]{\prod_{k=1}^n\frac{1}{x_k}}=n^2.$$
Also, the Tangent Line method helps.
Since our inequality is homogeneous, we can assume that $\sum\limits_{k=1}^nx_k=n$ and we need to prove that
$$\sum_{k=1}^n\frac{1}{x_k}\geq n$$ or
$$\sum_{k=1}^n\left(\frac{1}{x_k}-1\right)\geq0$$ or
$$ \sum_{k=1}^n\left(\frac{1-x_k}{x_k}+x_k-1\right)\geq0$$ or
$$\sum_{k=1}^n\frac{(x_k-1)^2}{x_k}\geq0$$ and we are done!
A: Jensen's inequality is $\rm{avg~}f(x_i) \le f(\rm{avg~} x)$ (for concave up functions).  So put in $f(z) = 1/z$:
$$\frac{1}{n} \sum_{i} \frac{1}{x_i} \le \frac{1}{\frac{1}{n} \sum_i x_i}$$
Note that this assumes $x_i > 0$.
