jmc

To begin, we first consider the $\frac{1}{\sqrt{3}+1}$ term. We can multiply both the numerator and the denominator by the conjugate of the denominator to get $$\frac{1}{\sqrt{3}+1}=\frac{1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{3-1}=\frac{\sqrt{3}-1}{2}.$$We can then substitute this back into our original expression and multiply both the numerator and denominator by $2$ to get $$\begin{array}{rl}\frac{1}{1+\frac{1}{\sqrt{3}+1}}& =\frac{1}{1+\frac{\sqrt{3}-1}{2}}\\ & =\frac{2}{2+\sqrt{3}-1}\\ & =\frac{2}{\sqrt{3}+1}.\end{array}$$If we multiply both the numerator and denominator of this expression by $\sqrt{3}-1$ and simplify, we end up with $$\begin{array}{rl}\frac{2}{\sqrt{3}+1}& =\frac{2}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}\\ & =\frac{2(\sqrt{3}-1)}{3-1}=\frac{2(\sqrt{3}-1)}{2}=\boxed{\sqrt{3}-1}.\end{array}$$