TITLE: Maximize an unweighted sum given a weighted sum
QUESTION [5 upvotes]: My problem boils down to the following:

Given real numbers $c_i \geq 0$
$$\begin{array}{ll} \text{maximize} & f(x_1)+f(x_2)+\cdots+f(x_n)\\ \text{subject to} & c_1x_1+c_2x_2+\cdots +c_nx_n = C\\ & x_i \geq 0\end{array}$$
It is furthermore known that  and $f$ is concave, strictly increasing and bounded.


I know that if we instead maximize $c_1f(x_1)+c_2f(x_2)+\ldots+c_2f(x_n)$, then we can apply Jensen's inequality and bound our sum from above by a constant and then apply the equality criterion
$$\frac{c_1f(x_1)+c_2f(x_2)+\ldots+c_2f(x_n)}{c_1+c_2+\ldots+c_n}\leq f\left(\frac{c_1x_1+c_2x_2+\ldots +c_nx_n}{c_1+c_2+\ldots+c_n}\right) = f\left(\frac{C}{c_1+c_2+\ldots+c_n}\right)$$
with equality iff 
$$x_1 = x_2 = \ldots = x_n = \frac{C}{c_1+c_2+\ldots+c_n}$$
Is there a way to salvage this solution for the unweighted case given weighted constraints?

REPLY [2 votes]: I think you are asking if you can get an "easy" solution that holds independent of the particular structure of $f$, that is, a solution that depends only on $C,c_1,...,c_n$.
Not in general. To see why, you can just do an example: Fix $C>0$. Fix $\epsilon>0$ and assume $c_i\geq \epsilon$ for all $i$.  Then any $(x_1, ..., x_n)$ that satisfies the constraints must have $x_i\leq C/\epsilon$ for all $i \in \{1, ..., n\}$. Fix $\alpha \in (0,1)$ and define
$$ f(x) =\left\{ \begin{array}{ll}
x^{\alpha} &\mbox{ if $0\leq x \leq C/\epsilon$} \\
(C/\epsilon)^{\alpha}  & \mbox{ if $x>C/\epsilon$} 
\end{array}
\right.$$
By a simple Lagrange multiplier argument you can show that the solution to maximizing $\sum_{i=1}^n f(x_i)$ subject to $\sum_{i=1}^n c_i x_i\leq C$ and $x_i\geq 0$ for all $i$ is: 
$$ x^*_i = \frac{Cc_i^{-1/(1-\alpha)}}{\sum_{j=1}^n c_j^{-\alpha/(1-\alpha)}} \quad \forall i \in \{1, ..., n\}$$
This solution requires knowledge of $\alpha$. 

On the other hand, the problem can easily be worked out by a Lagrange multiplier argument: 
Maximize 
$$\sum_{i=1}^n f(x_i) -\lambda \left(\sum_{i=1}^n c_i x_i - C\right)$$
over $x_i\geq 0$ for all $i \in \{1, ..., n\}$.  Then choose $\lambda$ to make the constraint $\sum_{i=1}^n c_ix_i = C$ hold.