# Augmented Lagrangian vs. Log Barrier

I would like to solve for $\max f(x_1,...,x_n)$ where $$f(x_1,...,x_n)=\frac{1}{\Big(1-\sum\limits_{i=1}^{n}x_i\Big)^{\big(1-\sum\limits_{i=1}^{n}x_i\big)}\prod\limits_{i=1}^{n}x_i^{x_i}}$$ for instance, if $n=3$, we have $$f(x_1,x_2,x_3)=\frac{1}{(1-x_1-x_2-x_3)^{1-x_1-x_2-x_3}x_1^{x_1}x_2^{x_2}x_3^{x_3}}$$

There is also the following constraint equations: $$c_0(x_1,...,x_n)=\sum\limits_{k=1}^{n}kx_k-1=0$$ and for $i=1,2,...,n$ we have $$c_i(x_i)=x_i<1$$ I am not sure if it is easier to use a log-barrier method or a augmented Lagrangian method, because the constraint is an inequality rather than an equality, where each variable $x_i$ is restricted to $[0,1]$.

Log-barrier method. Rewrite problem $$P:= \left\{ \begin{array}{rl} \max & f(x_1,\ldots,x_n) \\ {\rm s.t.} & 1\cdot x_1+2\cdot x_2+\ldots +n\cdot x_n=1\\ & 0\leq x_i\leq 1 \end{array} \right.$$ as problem $MP$ by adding variable $x_0=1-x_1-x_2-\ldots-x_n$ and $F(x_0,x_1,\ldots,x_n)=\frac{1}{x_0^{x_0}\cdot x_1^{x_1}\cdot x_2^{x_2}\cdot\; \cdots \;\cdot x_n^{x_n}}$ $$MP:= \left\{ \begin{array}{rl} \max & F(x_0,x_1,\ldots,x_n) \\ {\rm s.t.} & 0\cdot x_0+1\cdot x_1+2\cdot x_2+\ldots +n\cdot x_n=1\\ & x_0+x_1+x_2+\ldots+x_n=1\\ & 0\leq x_i\leq 1 \end{array} \right.$$ Let $\mu>0$ and a barrier $B(x_0,x_1,\ldots,x_n)=-\sum_{k=0}^{n}\Big(\log x_i + \log 1-x_i \big)$. The $\mu\!-\!MP$ problem is $$\mu\!-\!MP:= \left\{ \begin{array}{rl} \max & F(x_0,x_1,\ldots,x_n) +\mu B(x_0,x_1,\ldots,x_n) \\ {\rm s.t.} & 0\cdot x_0+1\cdot x_1+2\cdot x_2+\ldots +n\cdot x_n=1\\ & x_0+x_1+x_2+\ldots+x_n=1\\ & 0\leq x_i\leq 1 \\ & \mu>0 \end{array} \right.$$