The problem is to
minimize $ f(x_1, x_2 ,x_3, x_4):= - \Big[ \log ({\frac{1}{4} + x_1}) + \log ({\frac{1}{2} + x_2})+ \log ({\frac{1}{5} + x_3})+ \log ({\frac{3}{4} + x_4}) \big]$
such that $x_1 +x_2 +x_3 +x_4 =1, \quad x_i \geq 0$.
I formulated the Lagrangian $L(x, \lambda) = f(x_1, x_2 ,x_3, x_4) - \lambda(x_1 +x_2 +x_3 +x_4 -1)$.
Optimizing over the four variables respectively give $$ x_1 = -\frac{1}{\lambda} - \frac{1}{4}, \, x_2 = -\frac{1}{\lambda} - \frac{1}{2}, \, x_3 = -\frac{1}{\lambda} - \frac{1}{5}, \, x_4 = -\frac{1}{\lambda} - \frac{3}{4},$$ so $\lambda \geq - \frac{4}{3}$ ( for $x_4$ to be nonnegative). But plugging in to the feasible set gives $\lambda = - \frac{40}{27} < - \frac{4}{3}$. Hence, the Lagrange method does not work. Can we proceed as in LP, i.e. consider only the vertices of the feasible set, so at least one of $x_1, \ldots, x_4$ is zero?
If you're going to use a Lagrangian method you can't ignore the nonnegativity constraints. This is the correct Lagrangian: $$L(x,\lambda,z) = -\sum_i\log(x_i+b_i) - \lambda(\sum_i x_i - 1) - \sum_i z_i x_i$$ where $b=(1/4,1/2,1/5,3/4)$ and $z_i$ are the Lagrange multipliers for $x_i\geq 0$. These multipliers must be nonnegative. The optimality conditions are $$-\frac{1}{x_i+b_i} = \lambda + z_i, ~~ i=1,2,3,4, \quad \sum_i x_i = 1, \quad x_i\geq 0,~~i=1,2,3,4$$ Now here's a bit of a trick. Because $z_i\geq 0$, we can treat it like a slack variable and write those first equations as inequalities: $$-\frac{1}{x_i+b_i} \geq \lambda, ~~ i=1,2,3,4, \quad \sum_i x_i = 1, \quad x_i\geq 0,~~i=1,2,3,4$$ Solving for $x_i$ gives us $$ x_i \geq \max\{0, -\frac{1}{\lambda}-b_i\}, ~~ i=1,2,3,4, \quad \sum_i x_i = 1$$ Now, it takes a bit of a leap of faith, but if you assume equality here: $$ x_i = \max\{0, -\frac{1}{\lambda}-b_i\}, ~~ i=1,2,3,4, \quad \sum_i x_i = 1$$ what you can consider is sweeping $\lambda$ from $-\infty$ to $0$, and eventually you'll hit $\sum_i x_i = 1$. That's your answer! It's helpful to define $\sigma=-1/\lambda$, and sweep from $0$ to $+\infty$: $$ x_i = \max\{0, \sigma-b_i\}, ~~ i=1,2,3,4, \quad \sum_i x_i = 1$$ I found that $\sigma=0.65$, corresponding to $\lambda=-20/13$, gives me $x=(0.4,0.15,0.45,0)$. I can verify that this satisfies all of the optimality constraints in their original forms for appropriate choices of $z_i$, and I'm done.