I am having some trouble understanding a proof that uses arc length integration and I think I could use some help. Suppose we have a holomorphic function $f:\mathbb{D}\to\mathbb{C}$ that is 1-1 and $f(0)=0$. Suppose also that we have an inequality $|f'(z)|\geq\frac{1-|z|}{(1+|z|)^3}$ for all $z$ in the unit disk. My goal is to prove that if the segment $[0,f(z_0)]$ lies inside $f(\mathbb{D})$, then $|f(z_0)|\geq\frac{|z_0|}{(1+|z_0|)^2}$. This is part of the growth theorem from Duren's book.
Now let's suppose that we have the segment $[0, f(z_0)]$ and let $C$ be its preimage through $f$. Then obviously $f(z_0)=\displaystyle{\int_C f'(\zeta)d\zeta}$. Now it is $f(z_0)=\displaystyle{\int_C f'(\zeta)d\zeta=\int_a^b (f\circ C)'(t)dt}$. But as $t$ traverses $[a,b]$, $f\circ C$ traverses the segment that joins the origin with a point, hence the argument stays constant. Consequently,
$|f(z_0)|=\displaystyle{|\int_C f'(\zeta)d\zeta|=\int_C|f'(z)||dz|\geq\int_0^{|z_0|}\frac{1-s}{(1+s)^3}ds=\frac{|z_0|}{(1+|z_0|)^2}}$
My question concerns the inequality above. It is obvious in an intuitional way, but I can't get a rigorous proof. In my knowledge the arc length integral $\int_\gamma h(z)|dz|$ is defined as $\int_a^bh(\gamma(t))|\gamma'(t)|dt$. But if I use this definition in the inequality I get the following: $\displaystyle{\int_a^b |f'(C(t))||C'(t)|dt\geq\int_a^b\frac{1-|C(t)|}{(1+|C(t)|)^3}|C'(t)|dt}$.
But now I don't know how to handle this. Setting $u=|C(t)|$ is tempting but I believe this is not a $C^1$ diffeomorphism of the obvious intervals. Any ideas?
The following lemma should help:
Lemma: If $\gamma :[a,b]\to \mathbb C$ is differentiable, and $\gamma (t) \ne 0$ for $t\in (a,b],$ then
$$|\gamma |'(t) \le |\gamma'(t)|\,\,\, \text {for } t\in (a,b].$$
Proof: Let $\gamma (t) = x(t) + iy(t).$ Then $|\gamma (t)| = (x^2+y^2)^{1/2}.$ For $t\in (a,b],$ we have
$$|\gamma|' = \frac{xx' +yy'}{(x^2+y^2)^{1/2}} \le \frac{(x^2+y^2)^{1/2}((x')^2+(y')^2)^{1/2}}{(x^2+y^2)^{1/2}} = ((x')^2+(y')^2)^{1/2} =|\gamma '|.$$
Note that we've used Cauchy-Schwartz on the numerator.