The function $\sum_{j=1}^{N} w_{j}.\log y_{j}$ , subject to the constraints
$y_{j} \ge 0$ and $\sum_{j=1}^{N} y_{j} = 1$ , attains a global maximum at the single point
$$y_{j} = \frac{w_{j}}{\sum_{i=1}^{N} w_{i}} , j = 1,2,3, ..., N $$
How do you prove this?
$$f =\sum_{i=1}^n w_i\log\left(\frac{y_i}{w_i}\right) + \sum_{i=1}^n w_i\log\left(w_i\right)$$
The function $\log x$ is concave (as the second derivarive is equal to $-1/x^2$ and so negative). It suffices then to apply the Jensen's inequality:
$$\sum_{i=1}^n w_i\log\left(\frac{y_i}{w_i}\right)\le\left(\sum_{i=1}^n w_i \right)\cdot \log\left(\underbrace{\sum_{i=1}^n \left( w_i\cdot \frac{y_i}{w_i}\right)}_{=\sum_{i=1}^ny_i=1} /\left(\sum_{i=1}^n w_i\right)\right) =-\left(\sum_{i=1}^n w_i\right)\cdot\log \left(\sum_{i=1}^n w_i\right)$$ We deduce $$\color{red}{f \le -\left(\sum_{i=1}^n w_i\right)\cdot\log \left(\sum_{i=1}^n w_i\right) + \sum_{i=1}^n w_i\log\left(w_i\right)}$$
The equality occurs if and only if $\frac{y_1}{w_1}=...= \frac{y_n}{w_n} \iff y_i = w_i/ \sum_{j=1}^n w_j$ for $i=1,...,n$.