Problem Statement: Let $U\subset\mathbb{R}^{n}$ be open an convex, $\varphi:U\rightarrow \mathbb{R}^{n}$ of class $C^{1}$, with $\lVert d\varphi(x)\rVert \leq \lambda <1$. Then $f:U\rightarrow \mathbb{R}^{n}$ defined by $x\mapsto Ax+\varphi(x)$ ($A\in GL(\mathbb{R}^{n})$) is a diffeomorphism from $U$ onto $f(U)$.
I am having trouble with this proof. I have shown that $\varphi$ is a $\lambda$-contraction for $\lambda<1$. If I show that $\lambda<\frac{1}{\lVert A^{-1}\rVert}$, then it follows that $f$ is a homeomorphism (and form there I just need to show differentiability of $f$). My issue is showing that $\lambda<\frac{1}{\lVert A^{-1}\rVert}$.
I have tried using the triangle inequality on many different norm expressions, and the fact that $\varphi$ has a fixed point (since it is a contraction), and I have also tried using the fact that for any $x\in U$, $x=A^{-1}[f(x)-\varphi(x)]$.
Thus far, I have only been able to show that $$(1-\lambda)\lVert x-y\rVert\leq\lVert A^{-1}\rVert\lVert f(x)-f(y)\rVert,$$ $$\lVert A^{-1}\rVert\lVert f(x)-f(y)\rVert\leq (1+\lambda\lVert A^{-1}\rVert)\lVert x-y\rVert,$$ but I have not been able to construct an inequality helps me solve the problem. Is there a key fact that I am missing? Any suggestions are appreciated.