Let $C$ be the arc described by $z(t) = t+i(1-t^{3/2})^{2/3}$ for $0 \le t \le 1$. Show that $\frac{1}{Length(C)}\ \left| \int_{C}zdz\right| \le 2^{-1/6}$
A little stumped on how to continue on from a certain point I've reached. I've evaluated the integral $\left| \int_{C}zdz\right| = 1$ and found that the $Length(C) = \int_{0}^{1} \sqrt{t^2 + (1-t^{3/2})^2}$ which can't be evaluated as far as I'm concerned. So the inequality becomes $ 1 \le (2^{-1/6})(Length(C))$ and rearranging things, I want to now show that the $Length(C) \ge 2^{1/6}$, but I don't know how to show that this is true.
I found out its maximum is 1 at $t=0$. Just use the theorem about the bound of a contour integral and just find $\max|z|$, i.e. take the derivative and set to zero.