Function satisfying $(Df(x)h,h) \geq \alpha(h,h), \forall x,h \in \mathbb{R}^n$ has an inverse around every point?

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$f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ class $C^1$ such that: there exists $\alpha >0$:

$$(Df(x)h,h) \geq \alpha(h,h), \forall x,h \in \mathbb{R}^n$$

(where $(x,x)$ is the standard scalar product and $Df(x)$ is the differential of $f$ at point $x$)

Prove that the function satisfying that condition has an inverse around every point $x \in \mathbb{R}^n$.

My attempt;

To apply the inverse function theorem I would require that the determinant of the differential in canonical basis $e$ is not $0$.

I'm, however, not sure how to go about it. I only noticed that if $h = e_i$ you get that every diagonal element of the said matrix is bigger than $0$ but nothing more.

I'm not sure what more to do, I suppose I'm supposed to either prove that the matrix is symmetric, because then I would be able to conclude it's positive definite and therefore $det \neq 0$.