Let $\gamma(t) = e^{it}$ where $0 \le t \le 2\pi$. Prove that $$ \left|\int_{\gamma} \frac{1}{12+5z} dz \right| \le \frac{20}{91}\pi. $$
I tried ML inequality and got this integral $\le \frac{2\pi}{7}$. $$\left|\frac{1}{12+5z}\right| \le \left|\frac{1}{|12|-|5z|}\right| = \left|\frac{1}{12-5R}\right| = \frac{1}{7}$$ and length of the curve is $2\pi$.
How can I further improve the bound to be $\frac{20}{91}\pi$?
By Cauchy's Theorem the integral is $0$. The only pole, $-\frac {12} 5$, of the integrand is outside the contour since $|-\frac {12} 5|>1$.