Is this rational function convex?

Question

For $x_1, x_2 > 0$, let

$$f(x_1,x_2) := \frac{1}{x_1^3x_2}$$

Is function $f$ convex?

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I think it is because Hessian matrix is $$\begin{pmatrix} \frac{12}{x_1^5x_2} & \frac{3}{x_1^4x_2^2} \\ \frac{3}{x_1^4x_2^2} & \frac{2}{x_1^3x_2^3}\end{pmatrix}$$

And I think the matrix is positive because each value is positive (is it true?)

Thanks!

Answer 1

A twice differentiable function is convex iff its Hessian matrix is positive-semidefinite. Your Hessian matrix is good, but instead of considering its elements, you need to consider $$ \frac{12}{x_1^5x_2}x_1^2+\frac{3}{x_1^4x_2^2}(x_1+x_2)+\frac{2}{x_1^3x_2^3}x_2^2 $$ which also appears to be nonnegative given that $x_1,x_2>0$. And yes, it is convex.