Let $f(p)\in C^2(\mathbb{R}^n)$ be a function such that $\left(f_{p_i p_j}(p)\right)$ is positive definite for all $p\in \mathbb{R}^n$. It follows that
$$0<\lambda(p)\leq\Lambda(p),$$ where $\lambda(p),\Lambda(p)$ are the eigenvalues of $\left(f_{p_i p_j}(p)\right)$. For a function $v\in C^2(U), U$ open and bounded, the Euler operator is defined by
$$L(v):= f_{p_i p_j}(Dv) \frac{\partial^2 v}{\partial x_i \partial x_j}.$$
If $\lambda(p)\geq \nu \Lambda(p)$ holds for some $\nu>0$, I want to show that $L$ is uniformly elliptic: There exists a constant $\theta>0$ such that
$$f_{p_i p_j}(p) \zeta_i \zeta_j\geq \theta\lvert\zeta\rvert^2\quad \forall p,\zeta\in \mathbb{R}^n.$$
We have
$$f_{p_i p_j}(p) \zeta_i\zeta_j \geq \lambda(p) \lvert \zeta\rvert^2 \geq \nu \Lambda(p)\lvert \zeta\rvert^2$$
but I am struggling to find a lower bound for $\Lambda(p)$ for all $p\in \mathbb{R}^n$.
This statement appears for example in "Direct Methods in the Calculus of Variations" by Giusti on page 30.