Convexity of $\frac{1}{f}$ over the set where the concave function $f$ is positive $S \subset R^n,~~f : S \rightarrow R $ is a concave function.
$S^{'}= \{ x \in S: f(x)>0 \}. $  

Prove that $\frac{1}{f}$ is a convex function on $S^{'}.$

 A: Since concavity implies continuity, $S'$ is an open set and $\frac{1}{f}$ is a continuous function over $S'$, the convexity of $\frac{1}{f}$ follows from the midpoint-convexity, so we only have to prove
$$\frac{1}{f\left(\frac{x+y}{2}\right)}\leq\frac{1}{2}\left(\frac{1}{f(x)}+\frac{1}{f(y)}\right)\tag{1}$$
for any $x,y\in S'$ such that $\frac{x+y}{2}\in S'$ (we can remove this assumption since $S'$ is a convex set by the concavity of $f$). We can write $(1)$ as:
$$ f\left(\frac{x+y}{2}\right)\geq HM(f(x),f(y)), \tag{2}$$
but the concavity of $f$ over $S'$ gives:
$$ f\left(\frac{x+y}{2}\right)\geq AM(f(x),f(y)),\tag{3} $$
so the claim follows by the AM-HM inequality - equivalent to the convexity of the $\frac{1}{x}$ function over $\mathbb{R}^+$.

Provided that $S'$ is a convex set, another approach is to consider that $\log f$ is concave on $S'$ since it is the composition of two concave functions, then $-\log f$ is convex, then $\frac{1}{f}=\exp(-\log f)$ is convex since it is the composition of two convex functions.
A: I'll assume that $f$ is already positive, since you know that $S'$ is a convex set. So, $\log \circ f$ is concave: indeed, for $1/p+1/q=1$,
$$
\log f\left( \frac{x}{p}+\frac{y}{q} \right) \geq \frac{\log f(x)}{p} + \frac{\log f(y)}{q}
$$
is equivalent to
$$
\log f\left( \frac{x}{p}+\frac{y}{q} \right) \geq \log \left( f(x)^{\frac{1}{p}}f(y)^{\frac{1}{q}} \right),
$$
or equivalently
$$
f\left( \frac{x}{p}+\frac{y}{q} \right) \geq  f(x)^{\frac{1}{p}}f(y)^{\frac{1}{q}}.
$$
But 
$$
f(x)^{\frac{1}{p}}f(y)^{\frac{1}{q}} \leq \frac{f(x)}{p} + \frac{f(y)}{q}
$$
by some elementary inequality for positive numbers that you usually prove before Hölder's inequality. You conclude by the concavity of $f$.
Now, consider 
$$
\log \frac{1}{f} = -\log f,
$$
which is convex. Then $1/f = \exp \left( -\log f \right)$ is convex because the exponential is a convex increasing function.
