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 $[0,1]$. 
 A: 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 
$$
\mu\!-\!MP:=
\left\{
\begin{array}{rl}
\max       & F(x_0,x_1,\ldots,x_n) +\mu B(x_0,x_1,\ldots,x_n)                      \\
{\rm s.t.} & 0\cdot x_0+1\cdot x_1+2\cdot x_2+\ldots +n\cdot x_n=1\\
           & x_0+x_1+x_2+\ldots+x_n=1\\
           & 0\leq x_i\leq 1 \\
           & \mu>0                     
\end{array}
\right.
$$
