We have a positive definite quadratic form $Q(x,y)$, and we have a sequence $(x_n,y_n,z_n)$ satisfying $\lim_{n\to\infty}\frac1{z_n}Q(x_n,y_n) = 0$.
We want to show that $\lim\limits_{n\to\infty}\frac{x_n^3+\alpha y_n^3}{z_n^{3/2}}=0$, where $\alpha$ is a positive constant.
I've already shown it by setting $Q(x,y)=\epsilon$, using Lagrange multipliers to maximize $g_n(x,y)=\frac{x^3+\alpha y^3}{z_n^{3/2}}$, and obtaining a maximum value of $c\epsilon^{3/2}$ for some positive $c$. I suspect this is inelegant, and I wonder if there's some obvious better way that I'm missing.
I note that, in the calculation described above, I used the specific formula of $Q$, which I have available. I haven't provided it, because I'm looking for a solution that is independent of knowing that formula, because my intuition tells me it's something that should work for all positive definite quadratic forms. I hope there's a coordinate free argument, maybe working in the variables $\frac{x}{\sqrt{z}}$ and $\frac{y}{\sqrt{z}}$?
I'm grateful for any insights anyone can provide.
As $Q$ is positive-definite, there are $m$, $M>0$ with $$m(x^2+y^2)\le Q(x,y)\le M(x^2+y^2)$$ for all $x$ and $y$. Thus $$|x_n|^3\le(x_n^2)^{3/2}\le A Q(x_n,y_n)^{3/2}$$ for some constant independent of $n$. Likewise $$|y_n|^3\le Q(x_n,y_n)^{3/2}.$$ So $$\left|\frac{x_n^3+\alpha y_n^2}{z_n^{3/2}}\right| \le A(1+\alpha)\left( \frac{Q_n(x_n,y_n)}{z_n}\right)^{3/2}\to0.$$