I want to prove the following inequality, where $\gamma$ is a complex path with $[\gamma] \neq0$ and $f:[\gamma]\to \Bbb C$ a continuous function $$\biggl\lvert\int_\gamma \frac {f(z)}zdz\biggr\rvert^2<\biggl\lVert\frac 1 {z^2}\biggr\rVert_\gamma \cdot L(\gamma)\cdot\int_\gamma |f(z)|^2 dz$$
I tried proving it using the definition of the line integral and taking the absolute value of the integrand, so that I can use the Cauchy-Schwarz inequality. I looked for different approaches but always fail to get the integral on the right hand side. Instead I get $$\int_a^b |f(\gamma(t))|^2|\gamma'(t)| dt$$
which looks quite good for me, except for the absolute value of $\gamma'$.