This might be very trivial, but a proof I'm reading on the bounds of the zeta-function uses the following fact:
If $s=\sigma+it$ is a complex number and if $\sigma\geq 2$, then $|\zeta(s)|\geq 1-\displaystyle\sum_{n\geq 2}\frac{1}{n^2}$.
What would be a rigorous explanation of this? Many thanks in advance.
It's just the triangle inequality together with $\lvert n^{s}\rvert = n^{\operatorname{Re} s}$ and the monotonicity of $\sigma \mapsto n^\sigma$. Thus we have
\begin{align} \lvert \zeta(s)\rvert &= \Biggl\lvert 1 + \sum_{n = 2}^\infty \frac{1}{n^s}\Biggr\rvert\\ &\geqslant 1 - \Biggl\lvert \sum_{n = 2}^\infty \frac{1}{n^s}\Biggr\rvert\\ &\geqslant 1 - \sum_{n = 2}^\infty \frac{1}{\lvert n^s\rvert}\\ &= 1 - \sum_{n = 2}^\infty \frac{1}{n^\sigma}\\ &\geqslant 1 - \sum_{n = 2}^\infty \frac{1}{n^2}. \end{align}