Minimizing $\sum_{i=1}^n \frac{x_i^2}{w_i}$ subject to $\sum_{i=1}^n x_i=1$ 
Minimize $\displaystyle\sum_{i=1}^n \frac{x_i^2}{w_i}$ subject to $\displaystyle\sum_{i=1}^n x_i=1$.

The answer is $x_i=\displaystyle\frac{w_i}{\sum_i w_i}$ but I don't know why apart from plugging it in after finding the first derivative and setting to $0$. A hint is appreciated!
Edit: I get 
$$\begin{align}
\Lambda(v_j,\lambda) &= \sigma^2\sum\frac{v_j^2}{w_j}+\lambda\left(\sum v_j-1\right) \\
\frac{d}{dv_j}\Lambda(v_j,\lambda) &= 2\sigma^2\sum\frac{v_j}{w_j}+\lambda=0 \\
\frac{d}{d\lambda}\Lambda(v_j,\lambda) &= \sum v_j-1  = 0. \end{align}$$
Then $\lambda=-2\sigma^2\sum\frac{v_j}{w_j}$.
Not sure what's supposed to happen next.
 A: Without loss of generality $\sum\limits_{i=1}^nw_i=1$.
Now you can complete the squares like this
$$\sum_{i=1}^n\frac{x_i^2-2x_iw_i+w_i^2}{w_i}=\sum_{i=1}^n\frac{(x_i-w_i)^2}{w_i}$$
And note that $\sum\limits_{i=1}^n\frac{2x_iw_i}{w_i}=\sum\limits_{i=1}^n2x_i=2$ is a constant, so it's equivalent to minimizing this sum of squares.
So $x_i=w_i$ is the only minimum, because $\sum\limits_{i=1}^nx_i=\sum\limits_{i=1}^nw_i=1$ and if $x_i\ne w_i$ for any $i$, we can't have a minimum.
This is your answer as by assuming $\sum\limits_{i=1}^nw_i=1$ we just replaced each $w_i$ by $\frac{w_i}{\sum_{i=1}^nw_i}$.
A: Just to give an idea of how it would look using Lagrange multipliers: define $$f(x) = \sum_{i=1}^n \frac{x_i^2}{w_i} \quad \textrm{and} \quad g(x) = \big(\sum_{i=1}^n x_i\big) - 1$$
Then the goal is to minimize $f(x)$ subject to $g(x) = 0$. We now set $\enspace \nabla f = \lambda \nabla g \enspace$ and solve the resulting system of equations; in particular, this gives $\displaystyle \frac{2x_i}{w_i} = \lambda \enspace$ for $1 \leq i \leq n$. (If you haven't done Lagrange multipliers before, it's worth observing that while there are many $x_i$'s here, there is only one $\lambda$.)
Rearranging, we get $\displaystyle x_i = \frac{\lambda w_i}{2}$ for all $i$. Since $\lambda \neq 0$ (for then $x_i = 0$ for all $i$, making it impossible to satisfy $g(x)=0$), we immediately see that the $x_i$ are proportional to the $w_i$. The constraint $g(x) = 0$ allows you to find the appropriate value for $\lambda$.
Finally, you should note that we don't yet know whether we've found a maximum or a minimum, but I'll leave that as an exercise.
A: $
\def\b{\beta}\def\l{\lambda}\def\o{{\tt1}}\def\p{\partial}
\def\LR#1{\left(#1\right)}
\def\BR#1{\Big[#1\Big]}
\def\op#1{\operatorname{#1}}
\def\trace#1{\op{Tr}\LR{#1}}
\def\Diag#1{\op{Diag}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\c#1{\color{red}{#1}}
\def\CLR#1{\c{\LR{#1}}}
\def\fracLR#1#2{\LR{\frac{#1}{#2}}}
$For problems with simple constraints, a useful approach is to convert it into an unconstrained problem.
Introduce an unconstrained vector $u$ and define the variables
$$\eqalign{
x &= \frac{u}{\o:u} &\qiq \o:x=\frac{\o:u}{\o:u}=1 \\
v &= \o\oslash w &\qiq v\odot w=\o \\
V &= \Diag v  &\qiq Vw =\o \\
W &= \Diag w  &\qiq VW = I \\
\b &= v:\LR{x\odot x} \\
}$$
where $\{\,:\,\}$ denotes the Frobenius product, $\{\odot/\oslash\}$ denote the Hadamard product/quotient, $\o$ is the all-ones vector, and $\,\b\,$ is your cost function.
Calculate the gradient of $\b$ with respect to $u$ and solve for the optimal vector
$$\eqalign{
d\b &= v:\LR{2x\odot dx} \\
 &= \LR{2v\odot x}:{dx} \\
 &= 2Vx:\c{dx} \\
 &= 2Vx:\CLR{\frac{(\o:u)\,du-(\o:du)\,u}{(\o:u)^2}} \\
 &= \fracLR{2Vu}{\o:u}:\LR{\frac{(1:u)\,I-\c{u\o^T}}{(\o:u)^2}}du \\
 &= 2\LR{\frac{(\o:u)\,I-\c{\o u^T}}{(\o:u)^2}}\fracLR{Vu}{\o:u}:du \\
\grad{\b}{u}
 &= \fracLR{2V}{(\o:u)^3}\BR{(\o:u)\,u-\LR{u^TVu}w} \;\doteq\; 0 \\
u &= \fracLR{u^TVu}{\o:u}w \qiq \c{u_* = \l w} \\
}$$
Thus the optimal $u_*$ vector is a scalar multiple of $w$.
It is not necessary to determine the multiplier $\l$ to calculate the optimal $x$
$$\eqalign{
x_* &= \frac{u_*}{\o:u_*} = \frac{\l w}{\o:\l w} = \frac{w}{\o:w} \\
}$$
