prove this function: $f(x,y) = y^2/(x-x^2) $ with $x$ in $(0,1)$ is convex

Question

I want to show this function is convex. I know that this can be done by computing its hessian matrix. But this's gonna be tedious, that's why I would like to have a workaround.

Thanks in advance

Answer 1

$$\frac{\partial f}{\partial y} = \frac{2y}{x-x^2}$$ $$\frac{\partial^2 f}{\partial y^2} = \frac{2}{x-x^2} = \frac{2}{x(1-x)} > 0 ~\forall x \in \left(0,1\right)$$ $\therefore$ for $x \in \left(0,1\right) f$ is convex.

$$\frac{\partial f}{\partial x} = y^2 \cdot \frac{2x-1}{x^2(1-x)^2}$$ $$\frac{\partial^2 f}{\partial x^2} = y^2 \cdot \frac{-6x^2+6x-2}{x^3(1-x)^3}=y^2 \cdot \frac{-6\left(x-\frac{1}{2}\right)^2-\frac{1}{2}}{x^3(1-x)^3}>0~\forall y \in \mathbb{R}$$ $\therefore$ for $y \in \mathbb{R}$, $f$ is convex.

$\therefore f$ is convex $\forall x \in \left(0,1\right), y \in \mathbb{R}$.