Estimate of lower bound of $|s(s+1)|$ in contour integration

Question

I want to do following contour integration

$$\int_{c-i\infty}^{c+\infty} \frac {a^s}{s(s+1)} ds$$

Here c>0 ,a>1 and $\text{Re(s)=c}$ While doing a contour integration, I needed to estimate the lower bound of the function $\ |s(s+1)|$ where$\ s$ is complex number. The contour is shown in figure(below). Here$\ s$ is the complex number on the Semicircle $\text{C(T)}$ of given contour in figure,with the radius of semicircle being $\text{T}$. According to the book I have to prove: $$|s(s+1)| \geq T^2/2$$

I have been trying it for last 30 minutes but unable to prove it.

Edit: In the figure O is origin and C>0 is some point on real number line

Answer 1

$$\inf_{\Re(s) \le 0,|s|=T}|s(s+1)| = \inf_{\Re(s) \le 0,|s|=T}T |s+1|= T|T-1|$$ For $T$ very small and $T$ very large $T|T-1| \ge T^2/2$.

$T^2 - T=T^2/2$ means $T = 0$ or $T = 2$ thus $T|T-1| \ge T^2/2$ fails for $T \in [1,2)$ and it succeeds for $T \ge 2$.

And $T - T^2 = T^2/2$ means $T =0$ or $T = 2/3$ thus $T|T-1| \ge T^2/2$ fails for $T \in (2/3,1]$ and it succeeds for $T \in [0,2/3]$.