An inequality for an increasing, concave function

Question

Let $f:[0,\infty)\rightarrow[0,\infty)$ be a smooth function such that $f(0)=0$, $f''<0$, $f'(x)>0$ for $x\in(0,N)$ and $f'(N)=0$. I would like to show that $$ \max_{x\in(0,\infty)}\bigg(\sum_{k=1}^N \frac{1}{1+|f(k)-x|}\bigg)=\sum_{k=1}^N \frac{1}{1+f(N)-f(k)}, $$ i.e. the maximum is attained at $x=f(N)$.

The geometric considerations seem to confirm the validity of the above claim, but I don't know how to formalize the proof. Any hints will be appreciated!

Answer 1

I think $N = 3$, $f(x) = x \, (2 N - x)$ yields a counterexample. Denote $$ g(y) := \sum_{k = 1}^N \frac{1}{1 + |f(k) - y|}.$$

Then, $$ 1.1181 \approx g(f(2)) > g(f(3)) \approx 1.0928.$$

(Actually, one has to modify $f$ slightly to obtain a smooth function $f \colon [0,\infty) \to [0,\infty)$, but this can be done without changing $f(1), f(2)$ and $f(3)$)