Prove that if $|a| \neq R$ then $\int_{|z|=R} \frac{|dz|}{|z-a||z+a|} < \frac{2 \pi R}{|R^2-|a|^2|}$

Question

Prove that if $|a| \neq R$ then $$\int_{|z|=R} \frac{|dz|}{|z-a||z+a|} < \frac{2 \pi R}{|R^2-|a|^2|}$$

I am interested in knowing how to do this, particularly without any residue theory. Any hints are greatly appreciated! Thanks.

Answer 1

Recall that

$$\int_{|z|=R} |dz| = 2 \pi R$$

Now, by the triangle inequality:

$$\color{blue}{|z|^2-|a|^2} \leq |z^2-a^2| \leq \color{red}{|z-a||z+a|}$$

Hence,

$$\int_{|z|=R} \frac{|dz|}{\color{red}{|z-a||z+a|}} \leq \int_{|z|=R} \frac{|dz|}{\color{blue}{|z|^2-|a|^2}} = \frac{2\pi R}{|R^2-|a|^2|}$$

Still trying to figure out how to get the $\leq$ to become $<$ instead. Maybe this follows from the fact that $|a| \neq R$?