Prove convexity of $f(x_1,x_2,x_3)=\frac{x_1^2+x_1+1}{x_2+x_3}$ over $\{(x_1,x_2,x_3) : x_2,x_3\gt0\}$

Question

Prove convexity of $$f(x_1,x_2,x_3)=\frac{x_1^2+x_1+1}{x_2+x_3}$$ over $\{(x_1,x_2,x_3) : x_2,x_3\gt0\}$

I am looking for an easy (or relatively easy) way to show $f$ is convex. I tried to split the function into: $$\frac{x_1^2}{x_2+x_3}+\frac{1}{x_2+x_3}+\frac{x_1}{x_2+x_3}$$ The first and the second functions are convex:

• The first is convex because it's quadratic-over-linear which is convex;
• The second is convex because it's linear change of variables of $\frac{1}{x}$ which is convex over $x>0$.

I stuck proving the third one. I found it's Hessian matrix. Let $g(x)=\frac{x_1}{x_2+x_3}$.

$$\nabla ^2g(x)= \begin{bmatrix} 0 & \frac{-1}{(x_2+x_3)^2} & \frac{-1}{(x_2+x_3)^2} \\ \frac{-1}{(x_2+x_3)^2} & \frac{2x_1}{(x_2+x_3)^3} & \frac{2x_1}{(x_2+x_3)^3} \\ \frac{-1}{(x_2+x_3)^2} & \frac{2x_1}{(x_2+x_3)^3} & \frac{2x_1}{(x_2+x_3)^3} \end{bmatrix} $$

It suffices to show $\nabla g(x)\succeq0$, but I failed to do so. Last hope is calculating eigenvalues, but it seems too difficult. By the way, I can't use level-sets. Maybe there's an eaier way to show its convexity.

Please advise.

Thank you.

Answer 1

$g$ is not convex, as $(2,2)$-entry of $\nabla^2 g(x)$ is negative when $x_1$ is negative.

We have $\nabla f(x) = \begin{bmatrix} \frac{2x_1+1}{x_2+x_3} \\ - \frac{x_1^2+x_1+1}{(x_2+x_3)^2} \\ - \frac{x_1^2+x_1+1}{(x_2+x_3)^2}\end{bmatrix}$.

$$\nabla^2 f(x) = \begin{bmatrix} \frac{2}{x_2+x_3} & - \frac{2x_1+1}{(x_2+x_3)^2} & - \frac{2x_1+1}{(x_2+x_3)^2} \\ - \frac{2x_1+1}{(x_2+x_3)^2} & \frac{2(x_1^2+x_1+1)}{(x_2+x_3)^3} & \frac{2(x_1^2+x_1+1)}{(x_2+x_3)^3}\\ - \frac{2x_1+1}{(x_2+x_3)^2} & \frac{2(x_1^2+x_1+1)}{(x_2+x_3)^3} & \frac{2(x_1^2+x_1+1)}{(x_2+x_3)^3}\end{bmatrix}$$

One possible way to check positive semidefiniteness is by examine the determinant of each principal minor is nonnegative.

Since the second row and the third row is identitcal, the determinant is $0$. Also $|a_{23}|=0$.

Notice that $|a_{11}|>0$. Since $x_1^2+x_1+1 \ge \frac34> 0$, we have $|a_{22}|=|a_{33}|>0$.

$$|a_{12}|=|a_{13}|=\frac1{(x_2+x_3)^4}[4(x_1^2+x_1+1)-(2x_1+1)^2]=\frac{3}{(x_2+x_3)^4}>0.$$

Hence, it is convex.

Answer 2

Here is an easy approach:

We need that $g(x,y) = x^2/y$ is convex on $\mathbb R \times \mathbb R^+$, this follows e.g. from the theory of the 'perspective function'.

Then, we write \begin{align*} f(x_1, x_2, x_3) &= \frac{(x_1+1/2)^2 + 3/4}{x_2 + x_3} \\&= g(x_1 + 1/2, x_2 + x_3) + \frac{3/4}{x_2 + x_3}. \end{align*} Now, convexity follows easily.