If $\langle f'(x) \cdot v , v \rangle > 0$ then $f$ is injective

After a while I figured it out.

Consider the function $$\begin{align}\psi: [0,1] &\to \mathbb R\\ t &\mapsto \langle f(a + tv), v \rangle\end{align}$$

defined at $[0,1]$, differentiable with

$$\begin{align}\psi' (t) &= \langle f'(a + tv)\cdot v ,v\rangle + \langle f(a+tv),0 \rangle \\ &= \langle f'(a + tv)\cdot v,v \rangle > 0\end{align} \tag {*}$$

for $v = b -a$ and for all $t \in [0,1]$, then $\psi$ is strictly increasing at this closed interval. Now suppose that for $a \neq b$ we have $f(a) = f(b)$.

$$\begin{align}\psi (1) - \psi (0) &= \langle f(b), v \rangle - \langle f(a),v \rangle\\&= \langle f(b) - f(a), v \rangle\\ &= 0\end{align}$$

As $\psi$ is continuous and $\psi (1) = \psi (0)$ by Rolle's Theorem there exists $c \in (0,1)$ such that $\psi ' (c) = 0$, which contradicts $(*)$ .Thus $f(a) \neq f(b)$ and it follows that $f$ is injective.