Lagrange maximization with inequalities I need to prove the maxima of the following summation, using Lagrange. 
$$\max_{x_m} \left( \sum_m a_m log(x_m)\right) $$
s.t.
$$0 \le x_m \le 1$$
$$ \sum_m x_m = 1$$
The solution is a closed form $ x_m = \frac{a_m}{\sum_m a_m}  $ . 
I formulated the Lagrange equation but I am confused about the signs and the multipliers. 
$ L(x,\lambda,\mu) = \sum_m a_m log(x_m) + \sum_m \lambda_m (1-x_m) + \mu (\sum_mx_m-1) $ , Is this formulation correct ? what is wrong ?
note: only one $\mu$ for one constraint.

I think that the Lagrangian is:
\begin{equation}
L(x,\lambda,\mu) = \sum_m a_m log(x_m) + \mu \left(\sum_mx_m-1\right)
\end{equation}
Now we have:
\begin{equation}
   \begin{cases}
       \frac{a_1}{x_1}+\mu=0\\\frac{a_2}{x_2}+\mu=0\\\\\frac{a_m}{x_m}+\mu=0\\\sum_mx_m=1
   \end{cases}
\end{equation}
so 
\begin{equation}
   \begin{cases}
       x_1=-\frac{a_1}{\mu}\\x_2=-\frac{a_2}{\mu}\\\\x_m=-\frac{a_m}{\mu}\\\sum_mx_m=1
   \end{cases}
\end{equation}
and from the last equation: $-\frac{1}{\mu}\sum_ma_m=1$ so $\mu=-\sum_ma_m$. We can write the system like this:
\begin{equation}
   \begin{cases}
       x_1=-\frac{a_1}{-\sum_ma_m}\\x_2=-\frac{a_2}{-\sum_ma_m}\\\\x_m=-\frac{a_m}{-\sum_ma_m}\\\mu=-\sum_ma_m
   \end{cases}
\begin{cases}
       x_1=\frac{a_1}{\sum_ma_m}\\x_2=\frac{a_2}{\sum_ma_m}\\\\x_m=\frac{a_m}{\sum_ma_m}\\\mu=-\sum_ma_m
   \end{cases}
\end{equation}
