$M\mathcal{l}$-inequality: Putting a bound on the integrand in $\int_{|z|=R}\frac{f(z)}{(z-a)(z-b)}dz$

This question is related to another question here regarding an alternate proof of Liouville's Theorem using Cauchy's Integral Formula.

In trying to apply the $M\mathcal{l}-$inequality to the integral $\displaystyle \int_{|z|=R}\frac{f(z)}{(z-a)(z-b)}dz$, $\mathcal{l}$ obviously is $R$.

Since Liouville's Theorem assumes that $f(z)$ is bounded, we have that $|f(z)|\leq P$, so $\displaystyle \left\vert\int_{|z|=R}\frac{f(z)}{(z-a)(z-b)}dz \right\vert \leq \int_{|z|=R}\left\vert \frac{f(z)}{(z-a)(z-b)} dz\right\vert$. Now, we need this to be $\leq M\mathcal{l} = M\cdot R$.

However, I am having trouble figuring out what the denominator of our $M$ should be. I.e., it should be of the form $\frac{M}{N}$, where $\frac{1}{(z-a)(z-b)} \leq \frac {1}{N}$.

The answers to the problem I linked to seem to be under the impression that $N = R^{2}$. But, I think they got the inequality going the other way. I.e., since $|z-a|<R$, $|z-b|<R$, $\frac{1}{|z-a||z-b|} \geq \frac{1}{R^{2}}$, not $\leq \frac{1}{R^{2}}$. Maybe I'm misunderstanding what they're saying.

I have the rest of the proof done; I just need to bound the integral, so as $R \to \infty$, I'll have that the integral $\to 0$.
Hint: you are integrating over the circle $|z| = R$ and $$|z-a|\ge |z|-|a| = R-|a|,$$ $$|z-b|\ge |z|-|b| = R-|b|.$$
• @JessyCat ???? The line integral is a complex number and $0\le|\text{line integral}|$ is trivial. – Martín-Blas Pérez Pinilla Apr 5 '16 at 6:41