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The question is as follows. Let us consider a positive integer number $x \in \{1,2,3,...\}$ and a positive real number $q \in [1,x]$. Show that \begin{equation} \sum_{m=0}^{x-1} \left( \frac{x+2-q}{m+1} -\frac{x+q}{\min(m+q,x)} \right) \binom{x-1}{m} \left ( \frac{x-q}{x+q}\right)^m \geq 0. \end{equation}


I've tried to solve the inequality as follows. First, let us denote the LHS as $f(x,q)$
\begin{equation} f(x,q)=\sum_{m=0}^{x-1} \left( \frac{x+2-q}{m+1} -\frac{x+q}{\min(m+q,x)} \right) \binom{x-1}{m} \left ( \frac{x-q}{x+q}\right)^m. \end{equation} I want to show that $f(x,q) \geq 0$ for $x \in \{1,2,3,...\}$ and any positive real number $q \in [1,x]$. It is easy to verify that $f(x,1)=0$ and $f(x,x)=0$.

The first idea I pursued was to try by induction over $x$, while considering $f(x,q)$ with a fixed $q$. In fact, numerical results seem to verify that $f(x+1,q) \geq f(x,q)$ (the inequality is strict for $q > 1$). I tried to prove it by induction but I could not succed as I could not manage the terms $\binom{x-1}{m} \left ( \frac{x-q}{x+q}\right)^m$, which are very difficult to manipolate.

The second way I tried was to consider $f(x,q)$ by fixing $x$ and letting $q$ to vary in $[1,x]$. In this case $f(x,q)=0$ at the two extremes of the interval ($q=1$ and $q=x$), while numerical results show that $f(x,q)$ is greater than zero inside the interval, suggesting that $f(x,q)$ is quasiconvex in $q \in [1,x]$. My idea was to consider the different intervals of $q$ given by $q \in [1,2)$, $[2,3)$ (where $f(x,q)$ is differentiable), and show that in each of this interval the function is either convex or concave, while being quasiconcave all over the interval. In fact, by considering $q \in [a,a+1)$, for any integer $a$ in $[1,x]$, we have that
\begin{equation} f(x,q)=\sum_{m=0}^{x-a-1} \left( \frac{x+2-q}{m+1} -\frac{x+q}{m+q} \right) \binom{x-1}{m} \left ( \frac{x-q}{x+q}\right)^m + \sum_{m=x-a}^{x-1} \left( \frac{x+2-q}{m+1} -\frac{x+q}{x} \right) \binom{x-1}{m} \left ( \frac{x-q}{x+q}\right)^m . \end{equation} However, taking the derivate with respect to $q$ is very difficult and I wasn't able to conclude anything after this.

To sum up, it looks like that the issues are given by the terms $\binom{x-1}{m} \left ( \frac{x-q}{x+q}\right)^m$, which are very difficult to manage. If someone can help me by providing new directions to look at, pointing me out similar inequalities in the literature which I don't know, or give me new ideas, I would be grateful.

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After many attemps, I have simplified the inequality as follows: Show that, for any positive integer $x \in \{1,2,3,... \}$ and $q \in [1,x]$, the following inequality holds: \begin{equation} \sum_{m=1}^x \frac{m}{\min(m+q-1,x)} \binom{x}{m} \left( \frac{x-q}{x+q}\right)^m \leq \frac{x-q+2}{x+q} \left( \left(\frac{2x}{x+q} \right)^x -1\right). \end{equation} Note that the LHS and the RHS are equal for $q=1$ and $q=x$. Moreover, note that \begin{equation} \sum_{m=1}^x \binom{x}{m} \left( \frac{x-q}{x+q}\right)^m = \left(\frac{2x}{x+q} \right)^x -1 . \end{equation} This inequality looks more tractable. However, even here, all the classical techniques as CS, AM-GM, concavity/convexity seem to fail.

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