Cauchy-Schwarz and Jensen

Cauchy-Schwarz.
$$\left(\sum_{i=1}^n\sqrt{x_i}\cdot\frac{1}{\sqrt{x_i}}\right)^2\le\left(\sum x_i\right)\left(\sum\frac{1}{x_i}\right).$$
The LHS is $n^2$, so $n^2\le 1\cdot\sum\frac{1}{x_i}$.

Conclusion. $\sum\frac{1}{x_i}\ge n^2$, with equality iff all $x_i=1/n$.

Jensen approach. The function $f(x)=1/x$ is convex on $x>0$, so $\frac1n\sum f(x_i)\ge f(\bar x)=f(1/n)=n$, giving $\sum\frac{1}{x_i}\ge n^2$.