$a_1 + a_2 + \dots a_n = 1$ 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?

Consider the continuous function $f(a_1,a_2\dots,a_n)=a_1^2+a_2^2/2+\dots +a_n^2/n$. We will minimize it in the compact region defined by constraints $a_i\geq 0$ and $a_1+a_2+\dots + a_n=1$.
The idea is that since the region is compact the minimum must exist, and we will find necessary conditions for the point at which the minimum is achieved.
Suppose that $a_1+a_k=z$, then $a_1^2+\frac{a_k^2}{k}$ must be maximized with respect to $a_1+a_k=z$, when does this happen? we want to maximize $a_1^2+\frac{(z-a_1)^2}{k}$,the derivative of this function is $2a_1-\frac{2(z-a_1)}{k}$. So the minimum is reached when $(k+1)a_1=z$ or $a_1=\frac{z}{k+1}$, or when $ka_1=a_k$.
We conclude the maximum is reached at:
$a_1,2a_1,3a_1\dots na_1$, which clearly adds up to $1$ when $a_1=\frac{2}{n(n+1)}$.
