The Rosenbrock function is positive definite for all $x$ with $f(x) < 0.0025$

Question

Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ with $f(x) = 100(x_2-x_1^2)^2 + (1-x_1)^2$ be the Rosenbrock function.

Show that the Hessian of the Rosenbrock function is positive definite for all $x$ with $f(x) < 0.0025$

I thought I should use Sylverster's criterion here, so I calculated the Hessian of $f$:

$\nabla^2 f(x) = \begin{pmatrix} 1200x_1^2-400x_2+2 \quad& 400 \cdot x_1 \\ -400x_1 & 200\\ \end{pmatrix}$

and the determinants:

1. $det(\nabla^2 f(x)_1) = 1200x_1^2-400x_2+2$
2. $det(\nabla^2 f(x)_2) = 240000x_1^2-8000x_2+400+1600x_1^2$.

But I do not know how to proceed from here.

Could you help me?