Prove that if $\vert a \vert \neq R$,then $$\int_{\vert z\vert =R} \frac{\vert dz\vert}{\vert z-a\vert \vert z+a\vert}<\frac{2\pi R}{\vert R^2-\vert a\vert ^2\vert}...(1)$$
Proof:
We have the following result $\int_J \vert dz\vert=\int_a^b\vert J'(t)\vert dt=l(J)$ which is the length of $J$.
Then $\int_{\vert z\vert =R} \frac{\vert dz\vert}{\vert z-a\vert \vert z+a\vert}=\frac{1}{{\vert z-a\vert \vert z+a\vert}} \int _{\vert z\vert =R} \vert dz\vert=2\pi R\frac{1}{{\vert z-a\vert \vert z+a\vert}}$.
But $\frac{1}{{\vert z-a\vert \vert z+a\vert}}=\frac{1}{\vert R^2-\vert a^2\vert \vert}$.
Therefore $(1)$ would be an equality, where is my mistake?
Please be aware then when you write something like $$ \int_{0}^{a}f(z)\,dz = f(z)\int_{0}^{a} 1\,dz $$ a sweet kitten dies. To avoid a massacre, it is enough to notice that $$ \left|\oint_{|z|=R}\frac{dz}{z^2-a^2}\right| \leq \oint_{|z|=R}\frac{dz}{\left|z^2-a^2\right|} $$ by the triangle inequality, and by the triangle inequality $\left|z^2-a^2\right|\geq |z^2|-|a^2|$, hence: $$ \left|\oint_{|z|=R}\frac{dz}{z^2-a^2}\right| \leq \oint_{|z|=R}\frac{dz}{\left|z^2\right|-\left|a^2\right|} = \oint_{|z|=R}\frac{dz}{R^2-\left|a^2\right|}=\frac{2\pi R}{R^2-|a|^2}.$$