If $\|\left(f'(x)\right)^{-1}\|\le 1 \Longrightarrow$ $f$ is an diffeomorphism Let $f:\mathbb{R}^n \longrightarrow \mathbb{R}^n,f\in C^1(\mathbb{R}^n)$ such that $\forall x \in \mathbb{R}^n\;,\;f'(x)$ is an isomorphism and:
$$ \|\left(f'(x)\right)^{-1}\|\le 1\;,\forall x \in \mathbb{R}^n$$
How can we prove that  $f$ is a diffeomorphism ?  

Author's suggestion: 
$\forall [f(a),\color{red}b]\subset \mathbb{R}^n,\exists \lambda:[0,1]\longrightarrow \mathbb{R}^n$ such that $\lambda(0)=a$ and $f(\lambda(t))=(1-t)f(a)+t\color{red}b$
therefore $f$ is surjective.
Note that  $f$ is a local diffeomorphism, the strong part is to show that $f$ is injective.
There is a proof without using homotopy ?
Any hints would be appreciated.

Clearly the operator norm $ \|f'(x)\| \geq 1 $, hence $ f'(x)$ is invertible for all $ x \in \mathbb{R}^n $, from existence theorem of ODEs there exists $ \lambda : [0,1] \rightarrow \mathbb{R}^n $ satisfying
$$ \lambda'(t) = (f'(\lambda(t))^{-1}(b-f(a)),\ \ \ \lambda(0) = a $$
Hence we have the relation
$$ \begin{align}f(\lambda(t)) = f(\lambda(0)) + \int^t_0 (f\circ\lambda)'(s)ds\\ = f(a) + \int^t_0 f'(\lambda(s))(\lambda'(s))ds\\
=f(a) + \int^t_0 (b-f(a))ds= (1-t)f(a) +tb \end{align}$$
