Trying to prove $1/\sum_{i=1}^n \frac{a_i}{1-x_i}$ is (strictly) concave I am trying to show that the following function is (strictly) concave $$f(x_1,x_2,..,x_n)=\frac{1}{\frac{a_1}{1-x_1}+\frac{a_2}{1-x_2}+...+\frac{a_n}{1-x_n}} \text{, where } x_i\in (0,1)$$
Here $a_i$ are positive constants adding up to 1, that is $a_1+a_2+...+a_n=1$. The function is $n$-dimensional; so forming the Hessian and trying to show that it's negative definite proved futile. I can do it for $n=3$ but beyond that it becomes impractical.
I tried the following strategy, but got stuck at the end: The function $f$ is the reciprocal of $g$, where
$$g(x_1,x_2,..,x_n)=\frac{a_1}{1-x_1}+\frac{a_2}{1-x_2}+...+\frac{a_n}{1-x_n}$$ and it is easy to show that $g$ is an increasing and strictly convex function. Yet I could not prove that the reciprocal of such a function must be concave.
The functions look simple enough and I am pretty sure there is an easier way than hacking at the Hessian, but I just could not figure it out. 
Many thanks in advance...

Since there's no answer, let's follow Jonathan Davidson's suggestion. First of all, note that we can modify the problem to the equivalent problem of proving the concavity of,
$$f(x_1, \ldots, x_n) = \frac{1}{\frac{a_1}{x_1} + \ldots + \frac{a_n}{x_n}}$$
where $(x_1, \ldots x_n) \in (0, 1)^n$. It's also worth noting that the sum of $a_1, \ldots, a_n$ is immaterial, so long as they are all positive. At a point $(p_1, \ldots, p_n)$, we have the linearisation,
\begin{align*}
L(x_1, \ldots, x_n) &= f(p_1, \ldots, p_n) + \frac{\frac{a_1}{p_1^2}(x_1 - p_1) + \ldots + \frac{a_n}{p_n^2}(x_n - p_n)}{\left(\frac{a_1}{p_1} + \ldots + \frac{a_n}{p_n}\right)^2} \\
&= \frac{\frac{a_1}{p_1^2}x_1 + \ldots + \frac{a_n}{p_n^2}x_n}{\left(\frac{a_1}{p_1} + \ldots + \frac{a_n}{p_n}\right)^2}
\end{align*}
We need to show $f \le L$. Equivalently,
$$\left(\frac{a_1}{p_1} + \ldots + \frac{a_n}{p_n}\right)^2 \le \left(\frac{a_1}{p_1^2}x_1 + \ldots + \frac{a_n}{p_n^2}x_n\right)\left(\frac{a_1}{x_1} + \ldots + \frac{a_n}{x_n}\right).$$
If we replace $c_k = \sqrt{\frac{a_k}{p^2_k}x_k}$ and $d_k = \sqrt{\frac{a_k}{x_k}}$, then the inequality becomes,
$$(c_1 d_1 + \ldots + c_n d_n)^2 \le (c^2_1 + \ldots + c^2_n)(d^2_1 + \ldots + d^2_n),$$
which is just Cauchy-Schwarz. Hence $f$ is concave.
For your interest, if you have a function $g : \mathbb{R}^n \rightarrow (0, \infty)$, then $1/g$ being concave, implies $\log \circ g$ is convex, which implies $g$ is convex. Unfortunately, none of those implications are reversible!
