How to prove an inequality with positive numbers Let $n\in\mathbb N$ and $a_{1},...,a_{n} >0$, such that $a_{1} + \ldots +a_{n} =1$. Show that
$$
\left(\frac{1}{a_{1}^{2}} + \ldots +\frac{1}{a_{n}^{2}} \right)\left(2(a_{1}^{2}+\ldots+a_{n}^{2}) - \frac{1}{n}\right)\ge n^2.
$$
I tried induction, but it does not work.
Any help would be great.

Using Cauchy-Schwarz, we obtain
$$
a_1^2+\cdots+a_n^2\ge \frac{1}{n}(a_1+\cdots+a_n)=\frac{1}{n},
$$
and hence
$$
2(a_1^2+\cdots+a_n^2)-\frac{1}{n}\ge a_1^2+\cdots+a_n^2.
$$
Next, Cauchy-Schwarz again
$$
(a_1^2+\cdots+a_n^2)\left(\frac{1}{a_1^2}+\cdots+\frac{1}{a_n^2}\right)\ge (1+\cdots+1)^2=n^2,
$$
and hence
$$
\big(2(a_1^2+\cdots+a_n^2)-\frac{1}{n}\big)\left(\frac{1}{a_1^2}+\cdots+\frac{1}{a_n^2}\right)
\ge (a_1^2+\cdots+a_n^2)\left(\frac{1}{a_1^2}+\cdots+\frac{1}{a_n^2}\right)\
\ge n^2.
$$
