imc

For all nonzero numbers $a,$ recall that $a^{-1}=\frac{1}{a}$ is the reciprocal of $a.$ The original expression becomes $$\begin{array}{rl}{\left({\left((2+1{)}^{-1}+1\right)}^{-1}+1\right)}^{-1}+1& ={\left({\left(3^{-1}+1\right)}^{-1}+1\right)}^{-1}+1\\ & ={\left({\left(\frac{1}{3}+1\right)}^{-1}+1\right)}^{-1}+1\\ & ={\left({\left(\frac{4}{3}\right)}^{-1}+1\right)}^{-1}+1\\ & ={\left(\frac{3}{4}+1\right)}^{-1}+1\\ & ={\left(\frac{7}{4}\right)}^{-1}+1\\ & =\frac{4}{7}+1\\ & =\boxed{\frac{11}{7}}.\end{array}$$