To Show that the below complex integral vanishes
$$\int_{-\left(N+\frac{1}{2} \right)-i \infty}^{-\left(N+\frac{1}{2} \right)+i \infty}\Gamma(s) \,z^{-s}\,ds$$
Bruce C. Berndt uses the inequality
$$\bigg|\Gamma\left(\frac{1}{2}+it \right)\bigg|\geq N! \,\bigg|\Gamma\left(-N-\frac{1}{2}+it \right)\bigg|$$
He proceeds as following:
$$ \begin{aligned} \int_{-\left(N+\frac{1}{2} \right)-i \infty}^{-\left(N+\frac{1}{2} \right)+i \infty}\Gamma(s) \,z^{-s}\,ds&=O\left(\frac{1}{N!}\int_{-1}^{1}\bigg|\Gamma\left(\frac{1}{2}+it \right)\bigg|\,\big|z\big|^{N+\frac{1}{2}} \,dt \right)\\ & +O\left(\frac{1}{N!}\int_{1}^{\infty}e^{-\frac{1}{2}\pi t}\,\big|z\big|^{N+\frac{1}{2}}\,e^{t(\frac{\pi}{2}-\delta)} \,dt \right)\\ &=O\left(\frac{\big|z\big|^{N+\frac{1}{2}}}{N!} \right)+O\left(\frac{\big|z\big|^{N+\frac{1}{2}}}{N!} \int_1^\infty\,e^{-\delta t}\,dt\right)\\ &=O\left( \frac{\big|z\big|^{N+\frac{1}{2}}}{N!}\right)\\ &=O(1) \end{aligned} $$
I am trying to prove this inequality. By the Gamma´s function functional equation we can write
$$ \begin{aligned} \Gamma\left(\frac{1}{2}+it \right)&=\left(\frac{1}{2}-1+it \right)\Gamma\left(\frac{1}{2}-1+it \right)\\ &=\left(-\frac{1}{2}+it \right)\Gamma\left(-\frac{1}{2}+it \right)\\ &= \cdots \\ &=\left(-\frac{1}{2}+it \right)\left(-\frac{3}{2}+it \right) \cdots \left(-N-\frac{1}{2}+it \right)\Gamma\left(-N-\frac{1}{2}+it \right)\\ \end{aligned} $$
Taking modulus
$$ \begin{aligned} \bigg|\Gamma\left(\frac{1}{2}+it \right)\bigg|&=\bigg|\left(-\frac{1}{2}+it \right)\left(-\frac{3}{2}+it \right) \cdots \left(-N-\frac{1}{2}+it \right)\Gamma\left(-N-\frac{1}{2}+it \right)\bigg|\\ &=\bigg|\left(-\frac{1}{2}+it \right)\bigg|\bigg|\left(-\frac{3}{2}+it \right)\bigg| \cdots \bigg|\left(-N-\frac{1}{2}+it \right)\bigg|\bigg|\Gamma\left(-N-\frac{1}{2}+it \right)\bigg|\\ &=\sqrt{\frac{1}{4}+t^2 }\,\,\sqrt{\frac{9}{4}+t^2} \,\,\cdots \,\, \sqrt{N^2+N+\frac{1}{4}+t^2 }\,\,\bigg|\Gamma\left(-N-\frac{1}{2}+it \right)\bigg|\\ \end{aligned} $$
Now I got stuck in how to proceed
$$\sqrt{\frac{1}{4}+t^2 }\,\,\sqrt{\frac{9}{4}+t^2} \,\,\cdots \,\, \sqrt{N^2+N+\frac{1}{4}+t^2 }\geq 1 \cdot 2 \cdot 3 \cdots N $$
It doesn´t seem to make sense to me. Can someone give a Hint on how to finish the proof?