If derivative $f'$ of a function $f$ satisfies $0 < C \leq f'(x)$ for all $x$ then $f$ is bijective 
Proposition If derivative $f'$ of a function $f:\mathbb R\to\mathbb R$ satisfies $0 < C \leq f'(x)$ for all $x\in\mathbb R$ then $f$ is bijective.

It is clear that if there exist $a,b\in\mathbb R$ satisfy $f(a)=f(b)$ then there exist $c\in(a,b)$ such that $f'(c) = 0$ and this contradicts that $f'(x) > 0$, so necessarily $f$ is injective. However, I don't know how to prove that $f$ is surjective. I would be pleased if you give me a hint, thanks in advance.
 A: Hint: if $f$ is differentiable, it is continuous.  Then use the intermediate value theorem.
Hint, part 2: you might first show that $\lim_{\vert x\vert\rightarrow\infty}f(x)=\pm\infty$.
Showing the limit is a tiny bit trickier than I originally thought - we need to use the fact that $f^\prime\geq C>0$.  Suppose that $y>x$; then, by MVT,
$$
\frac{f(y)-f(x)}{y-x}\geq C>0
$$ and so 
$$
f(y)\geq f(x)+C(y-x)
$$ Now take an increasing, unbounded sequence (say, $x_n=n$).  The above tells us that 
$$
f(n+1)\geq f(n)+C\\
\ldots\\
f(n+k)\geq f(n)+kC\\
$$ Since $C>0$, it is now obvious that $\lim_{x\rightarrow\infty}f(x)=\infty$.
Similar proof for $x\rightarrow-\infty$.
A: We have $f'>0$ so $f$ is monotonically increasing.
If $t>0$ then 
$$\int_0^t f'(x)dx=f(t)-f(0)\geq \int_0^t Cdx=Ct$$
so we find 
$$\lim_{t\to+\infty}f(t)=+\infty$$
and 
If $t<0$ then 
$$\int_t^0 f'(x)dx=f(0)-f(t)\geq \int_t^0 Cdx=-Ct$$
so we find 
$$\lim_{t\to-\infty}f(t)=-\infty$$
now we can conclude.
Added Since the above proof need the assumption  that $f$ is locally integrable I add here another proof
If $t>0$ then by mean value theorem there's $\xi_t\in(0,t)$ such that 
$$f(t)-f(0)=f'(\xi_t)t>Ct$$
and then 
$$\lim_{t\to+\infty}f(t)=+\infty$$
If $t<0$ then by mean value theorem there's $\xi_t\in(t,0)$ such that 
$$f(t)-f(0)=f'(\xi_t)t<Ct$$
and then 
$$\lim_{t\to-\infty}f(t)=-\infty$$
