Proving that $|\zeta(\sigma+it)|^{-1}\leqslant 4$

Question

I am asked to prove that $|\zeta(\sigma+it)|^{-1}\leqslant 4$ for all $\sigma\geqslant 2$ and $|t|\geqslant 1$. Of course, one can use the fact that $$ \zeta(\sigma+it)^{-1}=\sum_{n=1}^{+\infty}\frac{\mu(n)}{n^{\sigma+it}} $$ and therefore $|\zeta(\sigma+it)|^{-1}\leqslant\zeta(\sigma)\leqslant\frac{\pi^2}{6}$. But this is not in the spirit of the question I am being asked as the Möbius function is not supposed to be known. Instead, I was first asked to show that $\zeta(\sigma)\leqslant 1+\frac{1}{2^{\sigma}}+\frac{1}{(\sigma-1)2^{\sigma-1}}$, which comes from an integral comparison. I don't know how to get the upper bound from this, any help is appreciated.