$a_1 + a_2 + \dots a_n = 1$ find min of $a_1^2 +\frac{a_2^2}{2} + \dots + \frac{a_n^2}n.$ Given $n$ numbers $a_1$, $\cdots$ and such that $a_1$, $a_2$, $\cdots$, $a_n > 0$ and their sum is $1$, I want to find the minimum value of 
$$a_1^2 + \frac{a_2^2}{2} + \cdots + \frac{a_n^2}{n}.$$
I have tried using weighted AM-GM inequality, like this:
$$\frac{a_1^2 + \frac{a_2^2}{2} + \cdots + \frac{a_n^2}{n}}{a_1 + a_2 + \cdots + a_n } \geqslant \frac{a_1^{a_1} \cdots a_n^{a_n}}{2^{a_2} \cdots n^{a_n}}$$
but was unable to make progress on the right hand side.
Is there a better way to apply AM-GM inequality? Or is there some different way altogether to solve this?

By the Cauchy-Schwarz inequality we have
$$ \left(\sum_{i=1}^{n}i\right)\left(\sum_{i=1}^{n}\frac{a_i^2}{i}\right) \geq \left(\sum_{i=1}^{n}\sqrt{i}\cdot\frac{a_i}{\sqrt{i}}\right)^2 = 1 \tag{1}$$
hence with our hypothesis we have:
$$ \sum_{i=1}^{n}\frac{a_i^2}{i}\geq \frac{2}{n(n+1)}\tag{2} $$
and equality is achieved iff $\frac{a_i}{\sqrt{i}}=\lambda\sqrt{i}$, i.e. iff $a_i = \frac{2i}{n(n+1)}$.
