Proving $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$. How would I show $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$?
I tried starting from the LHS, and rationalising and what-not but I can't get the result...
Also curious to how they got the LHS expression from considering the right.

By one line:$$\sqrt{3+\sqrt{13+4\sqrt{3}}} =\sqrt{3+\sqrt{1+2\cdot2\sqrt3+12}}= \sqrt{3+1+2\sqrt3}=1+\sqrt3$$ 
