About Finding Global Maximum and Proving its uniqueness The problem is as shown. I tried using gradient and Hessian but can not make any conclusions from them. Any ideas?
$$\max x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}$$
subject to
$$\sum_{i=1}^nx_i=1,\quad x_i\geq 0,\quad i=1,2,\ldots,n,$$
where $a_i$ are given positive scalars. Find a global maximum and show that it is unique.
 A: Calling $f = \prod_{k=1}^n x_k^{\alpha_k}$ we have the Lagrangian
$$
L(x,\mu) = f - \mu\left(\sum_{k=1}^n x_k-1\right)
$$
then the stationary points are determined by
$$
\alpha_k \frac{f}{x_k}-\mu = 0,\ \ \ k = 1,\cdots, n\\
\sum_{k=1}^n x_k-1=0
$$
or making $\lambda = \frac{\mu}{f}$
$$
\frac{\alpha_k}{x_k}-\lambda = 0\\
\sum_{k=1}^n x_k-1=0
$$
$\lambda$'s value is obtained substituting $x_k = \frac{\alpha_k}{\lambda}$ so
$$
\sum_{k=1}^n \alpha_k-\lambda=0
$$
and also $x_k = \frac{\alpha_k}{\sum_{k=1}^n \alpha_k}$
etc.
A: It looks like a homework, so I'll give you a hint instead of a complete solution. To maximize your function is the same as to maximize
$$
F(x)=\ln(x_1^{a_1}x_2^{a_2}\ldots x_n^{a_n})=\sum_{i=1}^na_i\ln x_i.
$$
The condition $x_i\ge 0$ cannot be active at the maximum, hence, we do not need it in the Lagrange equation for the gradients. Use Lagrange multipliers to conclude the necessary condition as
$$
\frac{a_i}{x_i}+v=0.
$$
Express each $x_i$, use the equality condition to find $v$, and finally find $x_i$.
The new problem is concave, then the Lagrange condition is also sufficient. The objective function $F$ is strictly concave, hence, the maximum is unique.
