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...
 A: 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!
