f is differentiable on convex domain D and Re(f')>0 implies f is injective on D

Suppose that $f$ is differentiable on a convex domain (open and connected) D and Re($f')>0$ in D. How can we prove that f is injective on D?

I found some answers using the mean value theorem for holomorphic functions, but I am not familiar with that theorem. Actually I want to use the mean value theorem for integrals, which tells us that the value of an differentiable function $f$ at the center $z_{0}$ of the disk $|z-z_{0}|<r$ is equal to the integral mean of its values on the boundary of that disk: $f(z_{0})=\int^{2\pi}_{0}{f(z_{0}+re^{it})dt}$.

Can you help me proving it by the mean value theorem for integrals?

I don't see how the mean value property of holomorphic functions - that $$f(z_0) = \frac{1}{2\pi}\int_0^{2\pi} f(z_0+re^{it})\,dt$$ for all $r$ small enough that the disk $\overline{D_r(z_0)}$ is contained in the domain of $f$ - can be employed here.

The result follows from an application of the fundamental theorem of calculus,

$$f(z_2) - f(z_1) = \int_0^1 \frac{d}{dt} f(z_1+t(z_2-z_1))\,dt = \int_0^1 f'(z_1+t(z_2-z_1))\,dt\cdot (z_2-z_1),$$

since $D$ is convex. The other premise yields that the last integral never vanishes.