Minimize $\sum a_i^2 \sigma^2$ subject to $\sum a_i = 1$ 
$$\min_{a_i}  \sum_{i=1}^{n} {a_i}^2 \sigma^2\text{ such that }\sum_{i=1}^{n}a_i=1$$ and $\sigma^2$ is a scalar.  

The answer is $a_i=\frac{1}{n}$.
I tried Lagrangian method. How can I get that answer?  

We have by Cauchy that sum $a_i^2$ times the sum $1$ is at least $1$, so the sum $a_i^2$ is at least $1/n$.  The problem is equivalent to minimizing the sum $a_i^2$ which happens when all the $a_i$ are equal by the equality case of Cauchy.
Hence, we do indeed have that $a_i=1/n$ minimizes it.
