There two definition of elliptic symbol.
A smooth matrix function $p(x,\xi)$ is a elliptic symbol of order $m\in\mathbb{R}$ if exist a constant $c>0$ such that for all $|\xi|>c$ we have $p(x,\xi)$ is invertible and
$\begin{align} (1) \hspace{3 cm} |p(x,\xi)^{-1}|\leq c (1+|\xi|)^{-m}. \end{align}$
And other definition is: a smooth matrix function $p(x,\xi)$ is a elliptic symbol of order $m\in\mathbb{R}$ if exist a constant $c>0$ such that for all $|\xi|>c$ we have
$\begin{align} (2) \hspace{3 cm} |p(x,\xi)|\geq c (1+|\xi|)^{m}. \end{align}$
How prove that both are equivalent?. is very easy show that (1) imply (2) because always $|p(x,\xi)|^{-1}\leq |p(x,\xi)^{-1}|$.
Any idea for show (2) imply (1). Thanks
I do not know what your definition of $|p(x,\xi)|$ is. If it is $$|p(x,\xi)| = |\det p(x, \xi) |$$ then the proof is easy. Indeed, (2) shows that $\det p(x, \xi) \neq 0$ for $|\xi|\geq c$ so the matrix $p(x, \xi)$ is invertible. Then, applying $\det$ in the equality $p(x, \xi) p(x, \xi)^{-1} = I$ gives $$|p(x,\xi)^{-1}| = \frac{1}{|p(x, \xi) |}.$$ Hence, $|p(x,\xi)^{-1}| \leq \frac{1}{C} (1 + |\xi|)^{-m}$ and you have (1) with $C^\prime = \max(C, \frac{1}{C})$.
If $|p(x,\xi)|$ is only a norm on the set of matrix, then the equivalence between your two definitions of ellipticity is not true. For example, if $|p(x,\xi)|$ is defined by $$|p(x,\xi)|^2 = \sum |p(x,\xi)_{ij}|^2$$ then take $p(x, \xi) = \mathrm{diag} ((1+|\xi)|)^m, 0, \cdots, 0)$ (regularized near $\xi = 0$) : this satisfies (2) but not (1) because it is not invertible.