Minimizing half concave half convex function

I am trying to prove the following theorem:

Theorem:
Let $ x_1, x_2, \ldots, x_n $ be real numbers satisfying the following conditions:

1. $ x_1 \leq x_2 \leq \cdots \leq x_n $,
2. $ x_1, x_2, \ldots, x_n \in [a, b] $
3. $x_1 + x_2 + \cdots + x_n = C $ (where $ C$ is a constant)

and let $ f: [a, b] \to \mathbb{R} $ be a function such that:

1. $ f $ is concave on $[a, c]$ and convex on $[c, b]$, for some $ c \in [a, b] $.

Define$$F = f(x_1) + f(x_2) + \cdots + f(x_n).$$Then: $ F $ is minimized when $ x_1 = x_2 = \cdots = x_{k-1} = a $, and $ x_{k+1} = \cdots = x_n $, for some $ k = 1, 2, \ldots, n $.

I have found the proof of it here: https://artofproblemsolving.com/community/c6h64933

However, I am unable to follow all the arguments. Specifically, I can't fully follow the induction step.

Question: I would appreciate a reference if this theorem is know or full proof is available.

Edit: From the link, I don't follow the following claim:

case(ii) assume $m,(1 \le m \le i)$ be the minimum interger such that $x_1+x_2+\cdots+x_m-(m-1)a \ge c$then we obtian$f(x_1)+f(x_2)+\cdots+f(x_m) \ge (m-1)f(a)+f(x_1+x_2+\cdots+x_{m-1}-(m-2)a)+f(x_m) \ge (m-1)f(a)+f(x_1+x_2+\cdots+x_{m-1}+x_m-c-(m-2)a)+f(c)$we reduce it to be $i-1$ case.so the theorem is proven.

Specifically, I fallow the first inequality but not the second. Why is that\begin{align*}&(m-1)f(a)+f(x_1+x_2+\cdots+x_{m-1}-(m-2)a)+f(x_m) \\&\ge (m-1)f(a)+f(x_1+x_2+\cdots+x_{m-1}+x_m-c-(m-2)a)+f(c)\end{align*}

I also don't follow the last line: "we reduce it to be $i-1$ case."