$\alpha_1, \ldots, \alpha_n$ are positive reals whose sum does not exceed one. It is required to prove that:
$$\sum_{i} \frac1{\alpha_i} \ge n^2$$
I would show my work, but I am certain that it does not offer any insight because I feel that there is just a trick that should be used, and I'm not seeing it.
Thank you.
Consider $x_i = \alpha_i^{\frac{1}{2}}$ and $y_i = \alpha_i^{-\frac{1}{2}}$ for $i = 1,2,\ldots,n$. By Cauchy-Schwarz inequality, $$\vert \langle x, y \rangle \vert \leq \|x\|_2\|y\|_2$$ which is $$\vert \sum_{i=1}^n x_i y_i \vert^2 \leq (\sum_{i=1}^n x_i^2)(\sum_{i=1}^n y_i^2)$$ and therefore $$n^2 = \vert \sum_{i=1}^n \alpha_i^{\frac{1}{2}}\alpha_i^{-\frac{1}{2}} \vert^2 \leq (\sum_{i=1}^n \alpha_i)(\sum_{i=1}^n \alpha_i^{-1}) = \sum_{i=1}^n \alpha_i^{-1}$$
By the arithmetic-harmonic mean inequality, we may conclude that $$\frac{\sum \alpha_i}{n}\ge \frac{n}{\sum 1/\alpha_i}$$ We rearrange this to get $$\sum \frac{1}{\alpha_i}\ge \frac{n^2}{\sum \alpha_i}$$ Since $\sum \alpha_i \le 1$, the desired inequality folows.
