If $a_i>0$ and $\sum_{i=1}^n a_i = 1$, then $\sum_{i=1}^n \frac{1}{a_i} \geq n^2$?

Question

If $a_i>0$ and $\sum_{i=1}^n a_i = 1$, is $\sum_{i=1}^n \frac{1}{a_i} \geq n^2$? I'm doing an inequality exercise. If I can confirm that's true, then my proof is done. I wrote down some examples and they are all true. I guess we need to compare each $\frac{1}{a_i}$ with $\frac{1}{n}$.

Answer 1

HM-AM says

$$\frac{n}{\displaystyle\sum_{k=1}^n\frac{1}{a_k}} \leq \frac{1}{n}\sum_{k=1}^na_k.$$

So, you are done.

Answer 2

Use that $$\frac{a_1+a_2+...+a_n}{n}\geq \frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}}$$ for all $$a_i>0,i=1...n$$

Answer 3

$$\sum_{i=1}^n\frac{1}{a_i}-n^2=\sum_{i=1}^n\left(\frac{1}{a_i}-n\right)=\sum_{i=1}^n\left(\frac{1-na_i}{a_i}+n(na_i-1)\right)=\sum_{i=1}^n\frac{(na_i-1)^2}{a_i}\geq0.$$