Please correct me where I am wrong in this prove of convexity?

Question

I have following function which I am trying to prove is convex $$f(\|\textbf{x}\|,y)=\frac{\|\textbf{x}\|^2}{y}$$ where $\textbf{x}$ is a two dimensional vector and $y\geq 0$. To show that $f(\|\textbf{x}\|,y)$ is convex I introduce a new vector $\textbf{z}=[\|\textbf{x}\|,y]$ and try to show that the following is true $$f(\lambda\textbf{z}_1+(1-\lambda)\textbf{z}_2)-\lambda f(\textbf{z}_1)-(1-\lambda)f(\textbf{z}_2)\leq0$$ where $\textbf{z}_i=[\|\textbf{x}\|_i,y_i]$. After doing some calculation I have following result for $f(\lambda\textbf{z}_1+(1-\lambda)\textbf{z}_2)-\lambda f(\textbf{z}_1)-(1-\lambda)f(\textbf{z}_2)$ $$-2y_1y_2\lambda^2\|\textbf{x}\|_1\|\textbf{x}\|_2-y_1y_2\|\textbf{x}\|_2^2-\lambda \left[(1-\lambda)y_2^2\|\textbf{x}\|_1^2+(1-\lambda)y_1^2\|\textbf{x}\|_2^2-2y_1y_2\|\textbf{x}\|_1\|\textbf{x}\|_2\right]$$ As we can see the first two terms are negative but how to show that the third term is also negative? Any help in this regard will be much appreciated. Please also point if this method of proving convexity is wrong. Thanks in advance.

Answer 1

You are asking if $f(u,y) = u^2/y$ is convex for $u,y \in [0,\infty)$. The Hessian at $(u,y)$ is $$\frac{2}{y}\begin{bmatrix} 1 & -u/y \\ -u/y & u^2/y^2\end{bmatrix}.$$ It suffices to check that this is positive semidefinite. This can be done by Sylvester's criterion (checking that all principal minors are nonnegative).