SETUP:

Assume $f(\cdot)$ is continuous and strictly monotone decreasing on $[0,\infty]$, with $f(0)>0$ and $f(\infty)<0$.

Let $x_m$ be the solution of $\frac{1}{m}\sum_{i=1}^{m}f(ix)=0$, where $m$ is some positive integer. (By the above assumption, the solution always exists and is unique.)

QUESTION:

Can someone give a necessary and sufficient condition such that there exists at least one $m$, such that $mx_m\geq(m+1)x_{m+1}$?

SOME COMMENTS:

This is a practical question I encounter in my work recently. It seems rather easy initially but I haven't been able to tackle it. I guess my maths is just deteriorating over years...

The question title might be a bit misleading. But lacking a better summary for the problem, I just put it that way. I apologize for the potential confusion.

For all "regular" functions I tried, the numerical result is always $mx_m<(m+1)x_{m+1}$. However, one can, for sure, construct weirdly shaped $f(\cdot)$ to achieve the reversed inequality. To give an example, let $f(x)$ be a piece-wise linear function that goes through points $(.37,.51)$, $(.5,.5)$, $(.74,.19)$, $(1,0)$, $(1.1,-.3)$, $(1.48,-.4)$, and $(1.5,-.5)$. It is so constructed that $x_3=.5$ and $x_4=.37$, which result in $3x_3=1.5>4x_4=1.48$. (Piece-wise linearity is not important here. One can always interpolate these points to get a sufficiently smooth, differentiable function.)

If easier, $f(\cdot)$ can be assumed to be differentiable on $[0,\infty]$. (But I doubt how much that will help...)

If necessary and sufficient condition is hard to derive, I am willing to see a sufficient condition that is not "too restrictive".

Well, thanks all for help in advance.