Ok, let's look at the denominator first. Good idea. So $1+\frac{1}{\sqrt{3}+1}=1+\frac{1(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}$. The denominator of that fraction is $(\sqrt{3}+1)(\sqrt{3}-1)=3-1=2$. The numerator is $1(\sqrt{3}-1)=\sqrt{3}-1$. Ok, so the whole fraction is $\frac{1}{\frac{\sqrt{3}+1}{2}}$. Which is equal to $2\div(\sqrt{3}+1)$. We get $\frac{2(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}$. The denominator is again 2. The numerator is $2(\sqrt{3}-1)=2\sqrt{3}-2$. And that simplifies to $\sqrt{3}-1$. And that is our final answer.


Simplify and rationalize the denominator: $$\frac{1}{1+ \frac{1}{\sqrt{3}+1}}.$$