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?  

$\displaystyle \sum_{i=1}^{n}(x-a_i)^2\ge0,\forall x\in \mathbb{R}$
$\displaystyle \Rightarrow \sum_{i=1}^{n}(x^2+a_i^2-2xa_i)\ge0$
$\displaystyle \Rightarrow nx^2+\sum_{i=1}^{n}a_i^2-2x\sum_{i=1}^{n}a_i\ge0$
Now we have a quadratic in $x$ which is always grater than equal to zero which implies that the quadratic can have either two equal roots in real nos. or has both complex roots.This implies that the discriminant is less than or equal to zero.
Discriminant $=\displaystyle D=4\left(\sum_{i=1}^{n}a_i\right)^2-4n\sum_{i=1}^{n}a_i^2\le0$
$\displaystyle \Rightarrow\left(\sum_{i=1}^{n}a_i\right)^2-n\sum_{i=1}^{n}a_i^2\le0$
$\displaystyle \Rightarrow 1-n\sum_{i=1}^{n}a_i^2\le 0$
$\displaystyle \Rightarrow \frac{1}{n}\le \sum_{i=1}^{n}a_i^2$
Equality holds if the equation has equal real root 
But then $\displaystyle \sum_{i=1}^{n}(x-a_i)^2=0$ for some $x\in R$
$\Rightarrow x=a_i,\forall 1\le i\le n$
$\Rightarrow \sum _{i=1}^{n}x=\sum _{i=1}^{n}a_i=1$
$\Rightarrow x=a_i=\frac{1}{n},\forall 1\le i\le n$
Now as $\sigma^2\ge 0$ so the min. of $\sum_{i=1}^{n}a_i^2\sigma^2$ is attained when $\sum_{i=1}^{n}a_i^2$ is also minimum.
I think this is a much better and elementary solution than solving it using Lagrange multipliers.
