Trying to solve next issue $$ \begin{cases} \sum_{i=1}^n x_i(ln_{x_i} - c_i) - min \\ \sum_{i=1}^n x_i = 1 \end{cases}$$ where $$ x_i, c_i \in R^n, c_i = const > 0 \forall i $$
I tried Lagrange method:
$$ L(x,\lambda) = \sum_{i=1}^n x_i(ln_{x_i} - c_i) + \lambda(\sum_{i=1}^n x_i - 1) $$
$$ \begin{cases} \ln x_i =c_i-(1+\lambda) \\ \sum_{i=1}^n x_i - 1 = 0 \end{cases} $$ so, $$ x_i = e^{c_i - (1+\lambda)} $$ and $\lambda$ gets from equation $$ \sum_{i=1}^n e^{c_i - (1+\lambda)} = 1 $$ Second derivative matrix is positive defined for all $x_i >0$ therefore this point is the solution.
Could you suggest me how to solve equation for $\lambda$ ?
From $\sum_{i=1}^n e^{c_i - (1+\lambda)} = 1$ we get $\frac{1}{e^{1+ \lambda}}\sum_{i=1}^n e^{c_i}=1$ thus $e^{1+ \lambda}=\sum_{i=1}^n e^{c_i}$, therefore
$$1 +\lambda= \ln (\sum_{i=1}^n e^{c_i}).$$