Conjecture about a property of concave functions

Question

Trying to prove a proposition in my paper, which can potentially use a conjecture about convex (concave) functions. This is likely to be wrong. I appreciate any thoughts on how to prove/disprove this.

Conjecture: For a concave function $f(x)\geq 0$, N is a positive integer, then $$ \frac{1}{N+1} \sum_{n=0}^N \frac{f(n)}{f(N)} $$ increases in N.

This statement is wrong for $f(x)=\sqrt{x}$. It is small disprove by counterexample. — Oleg567, Sep 14 '13 at 05:18
This is not true. For special cases like $f(x) = {x}^\theta $ I can prove this conjecture. Like what I did here — Chang, Sep 14 '13 at 13:50
Oh, I'm so sorry. It was mistake, because I considered expression $$ \frac{1}{N} \sum_{n=0}^N \frac{f(n)}{f(N)}.$$ — Oleg567, Sep 14 '13 at 18:41
I corrected my counterexample now: $f(x)=\sqrt{x+1}$ (in answer). — Oleg567, Sep 14 '13 at 19:07

score 0 · Accepted Answer · answered Sep 14 '13 at 19:06

Counterexample:

$$ f(x) = \sqrt{x+1}. $$

$N=1:~~~$ $\dfrac{1}{N+1}\sum\limits_{n=0}^{N}\dfrac{f(n)}{f(N)} = \dfrac{1}{2}\left(\dfrac{1+\sqrt{2}}{\sqrt{2}}\right) \approx 0.853553...;$

$N=2:~~~$ $\dfrac{1}{N+1}\sum\limits_{n=0}^{N}\dfrac{f(n)}{f(N)} = \dfrac{1}{3}\left(\dfrac{1+\sqrt{2}+\sqrt{3}}{\sqrt{3}}\right) \approx 0.797948...;$

$N=3:~~~$ $\dfrac{1}{N+1}\sum\limits_{n=0}^{N}\dfrac{f(n)}{f(N)} = \dfrac{1}{4}\left(\dfrac{1+\sqrt{2}+\sqrt{3}+2}{2}\right) \approx 0.768283...;$

$\cdots$

We can see that (at least) first terms are decreasing.

Conjecture about a property of concave functions

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