Given $f(x_1\ldots x_k) = \dfrac{x_1x_2\cdots x_k}{(x_0+c_1)(x_0+c_2)\cdots(x_0+c_k)}$
where $x_i > 0$, the $c_i > 0$ are constants, and
$$x_0 = \sum_{i=1}^k x_i$$
is it true that $f$ is quasiconcave? I suspect that it is, for the following reasons:
If $k = 2$, one can show directly using the bordered Hessian that $f$ is quasiconcave under the given conditions. Extending this to higher dimensions gets messy, though.
If all $c_i$ were set to the same value $c>0$, then $f$ is also quasiconcave. This is because $\log f(x) = \sum_{i=1}^k \log \frac{x_i}{x_0+c}$ can be seen to be an affine-fractional composition with a quasiconcave function, which is thus quasiconcave. This in turn implies that $f$ is quasiconcave because exponentiating is composition with an increasing function, which preserves quasiconcavity. Unfortunately, my $c_i$ are different, so this does not directly help.
I am a mere practitioner of convex optimization and no expert, so apologies for missing anything obvious.