Conditional Extremes/Lagrange multipliers: proving: $\frac{1}{x_1}+...+\frac{1}{x_n} \geq \frac{n^2}{x_1+...+x_n}$ $$\frac{1}{x_1}+...+\frac{1}{x_n} \geq \frac{n^2}{x_1+...+x_n};x_i>0.$$
This is supposed to be proven using conditional extremes. I tried the main function being $f(x_1,...,x_n)=\frac{1}{x_1}+...+\frac{1}{x_n} $and the condition:$x_1+...+x_n=s$ but when setting up the Lagrange equation I cannot get $x_i$ to be dependant upon $s$ and $n$ for reasons which are not clear. Help is appreciated.
 A: METHODOLOGY $1$:
As suggested in the comment from @EwanDelanoy, we can use the Cauchy-Schwartz Inequality when $x_i>0$ for all $i=1,\cdots,n$ by writing 
$$\begin{align}
n^2&=\left(\sum_{i=1}^n (1)\right)^2\\\\
&=\left(\sum_{i=1}^n \sqrt{x_i}\,\frac{1}{\sqrt{x_i}}\right)^2\\\\
&\le \left(\sum_{i=1}^n x_i\right)\left(\sum_{i=1}^n \frac{1}{x_i}\right)\\\\
&\frac{n^2}{\sum_{i=1}^n x_i}\le \sum_{i=1}^n \frac{1}{x_i}
\end{align}$$
as was to be shown!

METHODOLOGY $2$:
Let $f(x_1,x_2,\cdots,x_n)=\sum_{i=1}^n \frac{1}{x_i}-\frac{n^2}{\sum_{i=1}^n x_i}$.  Differentiating with respect to $x_j$ reveals and setting the derivative to $0$ reveals
$$\frac{1}{x_j^2}=\frac{n^2}{\left(\sum_{i=1}^nx_i\right)^2} \tag 1$$
for all $j=1,\cdots,n$.  From the $n$-equations in $(1)$, we find the solution for $x_i>0$ for all $i=1,\cdots,n$ is 
$$x_1=x_2=\cdots =x_n$$
Thus, the extreme value of $f$ occurs when all of the $x_i$ are the same at which $f=0$.  It is easy to show that this extreme value is a minimum (e.g., Let $x_i=1$ for $i\le n-1$.  Then, for any $x_n$, $f\ge 0$.) and hence we have
$$f(x_1,x_2,\cdots,x_n)\ge 0$$
And we are done!

METHODOLOGY $3$:
Note that 
$$\begin{align}
\sum_{i=1}^n \frac{1}{x_i}\,\sum_{j=1}^n x_j-n^2&=\sum_{i=1}^n \sum_{j=1}^n \left(\frac{x_j}{x_i} -1\right)\\\\
&=\frac12\sum_{i=1}^n \sum_{j=1}^n \left(\frac{x_j}{x_i}+ \frac{x_i}{x_j} -2\right)\\\\
&=\frac12\sum_{i=1}^n \sum_{j=1}^n \left(\frac{x_j^2-2x_ix_j+x_i^2}{x_jx_i}\right)\\\\
&=\frac12\sum_{i=1}^n \sum_{j=1}^n \left(\frac{(x_i-x_j)^2}{x_jx_i}\right)\\\\
&\ge 0
\end{align}$$
And we are done!
A: The inequality also follows directly from the fact that the arithmetic mean is greater than the harmonic mean $\frac{x_1+\ldots+x_n}{n}\geq \frac{n}{\frac{1}{x_1}+\ldots+\frac{1}{x_n}}$.
