Let $f: U \subset \mathbb{R}^n\to \mathbb{R}^m$ be a differentiable map in $U$, where $U$ is convex and open, such that $Df(x)$ is injective for all $x \in U$ and $$\langle Df(x)(x-y), Df(y)(x-y) \rangle >0$$ if $x\neq y$.
Show that $f$ is injective.
I don't see how one can prove the injectivity of $f$ by means of the mean value inequality as stated in the hyperlink. My approach would be the following.
Fix $x \ne y$, and put $h = y - x$. Consider the $C^1$ function $\varphi \colon [0,1] \to \mathbb{R}$ given by $$ \varphi(t) = \langle f(x + t h) - f(x), Df(x) h \rangle. $$ Its derivative equals $$ \varphi'(t) = \langle Df(x + t h) h, Df(x) h \rangle. $$ We have $\varphi(0) = 0$ and $\varphi'(0) = \langle Df(x) h, Df(x) h \rangle > 0$ (since $Df(x)$ is injective).
We claim that $\varphi'(t) > 0$ for any $t \in (0, 1]$. Indeed, we have $\langle Df(x + t h) h, Df(x) h \rangle > 0$ if and only if $\langle Df(x + t h) (t h), Df(x) (t h) \rangle > 0$, and the last inequality is just the assumption with $y$ replaced by $x + t h$.
Consequently, $\varphi$ is increasing on $[0, 1]$, hence $\varphi(1) > 0$, from which it follows that $f(y) \ne f(x)$.