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?  
 A: Use Lagrange multiplier as you tried.
$$ \begin{align*} 
f(a_1, \cdots, a_n) &= \sigma^2 \sum a_i^2 \\
g(a_1, \cdots, a_n) &= 1 - \sum a_i \\
F(a_1, \cdots, a_n; \lambda) &= f - \lambda g
\end{align*} $$
Partial derivatives are
$$ \begin{align*}
\frac{\partial F}{\partial a_j} &= 2 \sigma^2 a_j - \lambda \\
\frac{\partial F}{\partial \lambda} &= -1 + \sum a_i.
\end{align*} $$
From the constraint condition, $\frac{\partial g}{\partial a_i} = -a_i \neq 0$ for some $i$. So Lagrange multiplier guarantees those derivatives must be zero if $a_1, \cdots, a_n$ attain a minimum. From the $\frac{\partial F}{\partial a_j} = 0$, we get $a_j = \frac{\lambda}{2 \sigma^2}$. And from $\frac{\partial F}{\partial \lambda} = 0$, we get $\lambda = \frac{2 \sigma^2}{n}$ and hence $a_j = \frac{1}{n}$.
Let's check this gives us a minimum value indeed. If we "move" the point a little, that is, if we put $\alpha_i = \varepsilon_i + 1/n$ where $\sum \varepsilon_i = 0$, then it satisfies $g(\alpha_1, \cdots, \alpha_n) = 0$ but
$$ f(\alpha_1, \cdots, \alpha_n) = f\big(\frac{1}{n}, \cdots, \frac{1}{n}\big) + \sigma^2 \sum \varepsilon_i^2 \\
\geq f\big(\frac{1}{n}, \cdots, \frac{1}{n}\big) = \frac{\sigma^2}{n} $$
so $a_j = 1/n$ indeed attain a minimum.
A: I think you could approach this problem with the Cauchy-Schwarz inequality.
Using the Cauchy-Schwarz inequality, and the condition for it to be an equality you'll conclude that for the minimum value of $\sum{{a_i}^2 {\sigma}^2}$ is attained iff $a_i\sigma=\lambda$ for some $\lambda$ which is a constant. Since $\sigma$ is not equal to 0, this implies $a_i={\lambda}/{\sigma}^2$. This along with the constraint $\sum{a_i}=1$ yields $\lambda={{\sigma}^2}/n$ so that $a_i=1/n$
A: $\displaystyle \sum_{i=1}^{n}(x-a_i)^2\ge0,\forall x\in \mathbb{R}$
$\displaystyle \Rightarrow \sum_{i=1}^{n}(x^2+a_i^2-2xa_i)\ge0$
$\displaystyle \Rightarrow nx^2+\sum_{i=1}^{n}a_i^2-2x\sum_{i=1}^{n}a_i\ge0$
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.
A: 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.
