# Convex upper bound on a linear-fractional function

I have a function of the form $f(x,y) = \frac{x}{c+y}$ where $c$ is a positive constant, $c \ge x \ge 0$, and $y \ge 0$. I would like to find a convex upper-bound for this function. Is there a principled way for doing this? How about if the upper-bound has to be convex and piecewise linear? Is there a way to find the optimal upper-bound in terms of the number of linear pieces?

Update: Sorry I changed the boundary conditions on $x$ and $y$ so it suits my problem better.

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Let's first make sure that we speak the same language: "convex" means $f(\frac{a+b}2)\le\frac{f(a)+f(b)}2$. Are you sure that you mean this and not the opposite inequality, which I would normally call "concavity"? –  fedja Dec 6 '12 at 1:53
The function f is quasiliner. It has convex parts eg. when one sets $x=1$ and only varies $y$, $f(x,y) = f(y) = \frac{1}{c+y}$, and concave parts, eg. when one sets $y = x$, and so $f(x,y) = f(x) = \frac{x}{c+x}$. I would like to find a convex upper-bound for $f$ if possible. –  Norouzi Dec 6 '12 at 3:17

If you really mean $d\ge x$, then $f(x,y)\le f(d,y)$ which is convex, or even $f(x,y)\le f(d,0)$.
If it was a typo and the true domain is $[d,+\infty)\times[0,+\infty)$, the convex function $g(x,y):=f(x,0)+f(d,y)-f(d,0)=\frac{x}{c}+\frac{d}{c+y}-\frac{c}{d}$ has $g(d,y)=f(d,y)$ for all $y\ge 0$ and $g_x(x,y)\ge f_x(x,y)$ for all $x\ge d$ and $y\ge0$ , so $g\ge f$ on $[d,+\infty)\times[0,+\infty)$.
Not exactly the answer, but I think $g(x,y) = \frac{x^2}{(c+y)^2}$ is a reasonable convex lower-bound for $f$. With the assumption that $c \ge x$, then $1 \ge f(x,y) \ge 0$, and so $f(x,y)^2 \le f(x,y)$.