Problem statement
Consider the following rational function: $$ f(x) = -\frac{(x-a)(x-d)}{(x-b)(x-c)(x-e)} $$ where $a < b < c < d < e$.
I would like to prove that $f$ is convex on $(b,c)$.
Example
Here is an illustration of such a function with $(a,b,c,d,e) = (1,2,3,4,5)$.
Here is an interactive plot to visualize such a function made with geogebra:
https://www.geogebra.org/m/fcs8rbcr
What I tried
- I tried to calculate the second derivative which is quite complex.
- I tried to think in terms of variations of the function, we know its roots, its poles and its sign.
Perhaps there is a quick solution to such a problem.
The partial fraction decomposition of $f$ is $$ f(x) = \frac{B}{x-b} + \frac{C}{x-c} + \frac{E}{x-e} $$ with $$ B = - \frac{(b-c)(b-d)}{(b-a)(b-e)} > 0\\ C = - \frac{(c-a)(c-d)}{(c-b)(c-e)} < 0\\ E = - \frac{(e-a)(e-d)}{(e-b)(e-c)} < 0 $$ and on $(b, c)$ is $1/(x-b)$ convex, whereas $1/(x-c)$ and $1/(x-e)$ are concave.