Prove that, if $g(x)$ is concave, for $S = {x : g(x) > 0}$, $f(x) = 1/g(x)$ is convex over $S$. Using the definitions of convexity and concavity, I need to show the following:
$$g(ax + (1-a)y)\geq ag(x) + (1-a)g(y),\ \ a \in (0, 1)$$
implies that 
$$f(ax + (1-a)y) \leq af(x) + (1-a)f(y)\ , a \in (0, 1)\ .$$
I have tried every algebraic manipulation I can come up with to no avail. 
If I were to assume $f$ and $g$ are both twice differentiable, it is easy to show the second partial derivative of $f(x)$ w.r.t. $x$ is non negative, and hence, $f(x)$ is convex. However, I would like to prove it using the formal definition of convexity.
 A: Here is one way to do it:
$$\begin{align}
1 &= (a+(1-a))^2 \\
&= a^2 + 2a(1-a) + (1-a)^2 \\
&\le a^2 + a(1-a)\left(\frac{g(x)}{g(y)} + \frac{g(y)}{g(x)}\right) + (1-a)^2 \\
&= (ag(x)+(1-a)g(y)) \left(\frac{a}{g(x)} + \frac{1-a}{g(y)}\right) \\
&\le g(ax+(1-a)y) \left(\frac{a}{g(x)} + \frac{1-a}{g(y)}\right) \\
&= \frac{af(x) + (1-a)f(y)}{f(ax+(1-a)y)}
\end{align}$$
The first inequality is justified by the fact that $2 \le \frac{g(x)}{g(y)} + \frac{g(y)}{g(x)}$. You can prove this by using the AM-GM inequality.
A: Let  $0 < a < b < 1$. By concavity of $g(x)$, the graph of $g(x)$ between $(a,g(a))$ and $(b,g(b))$ lies on or above the line segment connecting $(a,g(a))$ and $(b,g(b))$. Since $g(a), g(b) > 0$, this line segment lies above the $x$-axis. 
Write the equation of this line as $y(x) = cx + d$. Taking reciprocals, we see that 
$${d^2 \over dx^2} \bigg({1 \over y(x)}\bigg) = {2c^2 \over (cx + d)^3}$$
Since $y(x) > 0$ between $x = a$ and $x = b$, we conclude that for $a < x < b$ one has
$${d^2 \over dx^2} \bigg({1 \over y(x)}\bigg) > 0$$
In other words, ${1 \over y(x)}$ is convex between $x = a$ and $x = b$. So the  graph of ${1 \over y(x)}$ lies 
on or below the line segment connecting $(a, f(a))$ and $(b, f(b))$. 
Since the graph of $g(x)$ between $(a,g(a))$ and $(b,g(b))$ lies on or above the line segment connecting $(a,g(a))$ and $(b,g(b))$, the graph of $f(x) = {1 \over g(x)}$ between $(a, f(a))$ and $(b, f(b))$ lies on or below the portion of the graph of ${1 \over y(x)}$ between $x =a$ and $x = b$, which we saw lies 
on or below the line segment connecting $(a, f(a))$ and $(b, f(b))$. So the graph of $f(x) = {1 \over g(x)}$ between $(a, f(a))$ and $(b, f(b))$ lies on or below the line segment connecting $(a, f(a))$ and $(b, f(b))$. Hence $f(x)$ is convex.
