Convexity of a rational function

Question

I am attempting to (dis)prove that the function

$$(x,y) \mapsto \frac{4x+3y+2}{x^2+xy+2x+y}$$

is convex for $x,y>0$.

Attempting to differentiate the function does not seem like a good idea (or am I making a mistake and this simplifies greatly?). I tried to search for useful rules of quotient of convex functions, however that did not yield and pertaining and/or useful information.

My best guess is I could somehow build an argument starting from the fact that the denominator is always positive.

Thanks for any help!

Answer 1

I guess you are right. Look at

http://www.stanford.edu/~boyd/cvxbook/

examples 3.32 and 3.33, you may follows these ideas.

Answer 2

Since no-one has answered this way or that...

The function is convex on $(0,\infty)^2$.

The direct brute-force approach (via Mathematica) considering derivatives works out fine:

Assume $x$ and $y$ are positive. The Hessian of the function at $(x,y)$ is

$$ \frac 1 {p(x,y)^3} \left( \begin{array}{cc} 8 x^3+18 x^2y+12 x^2+18 xy^2 +24 xy+24 x+6 y^3+14 y^2+20 y+16 & 2 x^3-2 xy +8 x+4 y+8 \\ 2 x^3-2 xy+8 x+4 y+8 & 2 x^3+2 x^2+4 x+4 \\ \end{array} \right). $$ Here, $p(x,y) = x^2+x y+2 x+y$ is the denominator of the original function. Since it's positive, we can disregard this factor and consider the matrix without it, which we'll call $H$.

Considering the following equation for the eigenvalues of $H$:

$$\frac {\text{tr}(H)\pm \sqrt{\text{tr}(H)^2 - 4\det(H)}}2$$

(where $\text{tr}(H)$ is the trace of $H$), it's sufficient for positivity of the eigenvalues that

• $\text{tr}(H) > 0$, and
• $\det(H) > 0$.

The first is evident from the above matrix, since all coefficients of the polynomials on the diagonal are positive. Regarding the second, the determinant works out to be

$$\det(H) = 12 x^6+36 x^5 y+40 x^5+36 x^4 y^2+92 x^4 y+72 x^4+12 x^3 y^3+64 x^3 y^2+144 x^3 y+128 x^3+12 x^2 y^3+96 x^2 y^2+240 x^2 y+112 x^2+24 x y^3+144 x y^2+144 x y+32 x+24 y^3+40 y^2+16 y,$$

which is again a polynomial with all-positive coefficients.

That is, both eigenvalues of the Hessian are positive for all positive $x$ and $y$, hence the function is convex.

So in some sense, it does "simplify greatly" once the algebra handle is cranked. I'm sorry I don't know a more elegant approach!

Remark 1: The same method applies to show the function is log-convex, a stronger property.

Remark 2: While considering this in Mathematica, I first accidentally put $p(x,y) = x^2+x y+2 x+y{}^2$ as the denominator, in which case the function is convex almost nowhere in $(0,\infty)^2$! So the underlying reason for convexity seems to be somewhat subtle. It would be interesting to construct similar examples.