Lagrange maximization with inequalities

Question

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.

Answer 1

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}