The question I'm struggling with is as follows:
Let $a>0$ and $\delta > 0$ (fixed). Suppose $ a \mapsto \frac{(a+\delta)_{m}}{(a)_{m}}$ is decreasing then prove that $ a \mapsto (a+\delta)_{m} - (a)_{m} $ is increasing ( means Wright convex).
I'm reading a paper in which the authors have given no reason for this. I have tried by taking $b>a$ and tried to show that $(a+\delta)_{m} - (a)_{m} < (b+\delta)_{m} - (b)_{m} $ using $\frac{(a+\delta)_{m}}{(a)_{m}} > \frac{(b+\delta)_{m}}{(b)_{m}}$ (decreasing). I have getting $(a+\delta)_{m} - (a)_{m} - (b+\delta)_{m} + (b)_{m} > \text{something} $. I can't get the answer to this. I guess there is something related to increasing and decreasing functions. I'm not getting an idea how to proceed.
Assuming $m>0$, suppose $x>0$ and $\delta>0$. Define $$h(x)= \frac{(x+\delta)_m}{(x)_m}.$$ Let, for the sake of convenience, $f(x) = (x+\delta)_m$ and $g(x) = (x)_m$. I now see that there is an extra information in your problem statement which was unnecessary. I see that with $\delta>0$, $h(x)$ is decreasing for $x>0$. To see this, logarithmically differentiate $h(x)$ to get,
$$h^{\prime}(x) = -\delta h(x) \left(\frac{1}{x(x+\delta)}+\dots+\frac{1}{(x+\delta+n-1)(x+n-1)}\right).$$
$h(x)$ is anyway positive. So, $h^{\prime}(x)$ is negative and hence $h$ is decreasing.
Quite clearly
$$f^{\prime}(x) = \frac{f(x)}{x+\delta}+\frac{f(x)}{x+\delta+1}+\dots+\frac{f(x)}{x+\delta+n-1}.$$
And
$$g^{\prime}(x) = \frac{g(x)}{x}+\frac{g(x)}{x+1}+\dots+\frac{g(x)}{x+n-1}.$$
Further, it is easy to see by the definitions of $f$ and $g$ that
$$\frac{f(x)}{x+\delta+k}-\frac{g(x)}{x+k} \geq 0,$$ whenever $k\geq 0$.
Thus
$$f^{\prime}(x) \geq g^{\prime}(x),$$ indicating that $f(x)-g(x)$ is increasing.