A proof of Popoviciu's inequality follows from Jensen's Inequality for convex functions. Wlog, assume $x \le y \le z$ there are two possibilities. Either $\displaystyle y < \frac{x+y+z}{3}$ or the reverse.
In the first case, $\displaystyle \frac{x+y+z}{3} < \frac{x+z}{2}< z$ and $\displaystyle \frac{x+y+z}{3} < \frac{y+z}{2} < z$, thus, there are positive real numbers $\lambda_1,\lambda_2 >0$ such that,
$\displaystyle \frac{x+z}{2} = \lambda_1 z + (1-\lambda_1)\left(\frac{x+y+z}{3}\right)$ and $\displaystyle \frac{y+z}{2} = \lambda_2 z + (1-\lambda_2)\left(\frac{x+y+z}{3}\right)$
It is easy to verify that $\lambda_1 + \lambda_2 = \dfrac{1}{2}$.
Now, using Jensen's inequality one has:
$\displaystyle f\left(\frac{x+y}{2}\right) \le \frac{f(x)+f(y)}{2}$
$\displaystyle f\left(\frac{x+z}{2}\right) = \lambda_1 f(z) + (1-\lambda_1)f\left(\frac{x+y+z}{3}\right)$
$\displaystyle f\left(\frac{y+z}{2}\right) = \lambda_2 f(z) + (1-\lambda_2)f\left(\frac{x+y+z}{3}\right)$
Adding the three inequalities gives the Popoviciu's inequality. The second case where $\displaystyle y > \frac{x+y+z}{3}$ we simply interchange the roles of $x$ and $z$ and the rest is similar.
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.
Best Answer
One way equality can occur is if $x_1 = \cdots = x_n$. Another way the equality can occur is if $f$ is linear.