prove $-\frac{c' x}{c(x)}$ is decreasing. falling elasticity. Something was asserted without proof in an economics paper but I would like to see it proven.
Suppose
$$
c(x) = p + v\cdot t(x),
$$
where $t>0$, $t'<0$, $t''>0$, $p>0$, $v>0$ and $c\rightarrow \infty$ as $x\rightarrow 0$.
It's claimed that
$$
-\frac{c'(x)x}{c(x)}
$$
is positive but decreasing---i.e., that the elasticity of $c(x)$ is falling in magnitude. Obviously it is positive (since $c'<0$) but I don't know how to prove it's decreasing.
I guess that means
$$
\frac{d}{dx} \frac{c'\cdot x}{c} = \frac{1}{c}\left[
c'' x + c' - \frac{(c')^2x}{c}
\right]=
\frac{1}{p+v t(x)}\left[
vt''(x) x + vt'(x) - \frac{(vt'(x))^2x}{p+vt(x)}
\right]
>0.
$$
I am unable to prove this to be so.
 A: As a partial result, it is true in the case of $t(x)=x^{-k}$, $k>0$. In this case:
$$
-\frac{c'(x) x}{c(x)} = -\frac{-kx^{-k-1}x}{p+vx^{-k}} = \frac{k}{v+px^k} 
$$
which is monotonically decreasing. It should also work for finite sums of such functions with non-negative coefficients. However, I also have my doubts whether it is true generally. We need $(c''c - (c')^2)x + cc' > 0$, so in particular already $c''c -(c')^2 > 0$ which would imply that $c$ is strictly logarithmically convex. So any convex, but not logarithmically convex function with the requested properties will be a counter-example, if such a function exists.
EDIT: This related question Is a decreasing and assymptotic (for y=const) convex function always logarithmically convex? guided me to a full counter example:
If $t(x) =\exp(-x)$ then $-\frac{c'(x) x}{c(x)} = \frac{x}{v+pe^x}$ which is increasing around $x=0$. To ensure that we have the requested pole at $x=0$ we can add a $\frac{1}{x}$ term - the example still works after some tweaking:
$$\begin{aligned} t(x)=\tfrac{1}{x} + \lambda e^{-x}
\implies
-\frac{c'(x) x}{c(x)}
=\frac{v\frac{1}{x}+ v\lambda x e^{-x}}{v(\frac{1}{x}+\lambda e^{-x}) + p}
= \frac{1 + \lambda x^2 e^{-x}}{1+\frac{p}{v}x+\lambda xe^{-x}}
\end{aligned}$$
Note that at $x=0$, the value is $1$ and at $x=\infty$ it is $0$. By taking $\lambda$ large enough (depending on $v$ and $p$), we can always enforce the existence of a local maximum in between, since then the $x^2e^{-x}$ term will dominate for some interval. Hence, the function cannot be monotonically decreasing. I guess it is time to contact the authors of the paper.
