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.

We have $$\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3};$$ that is,
$$3+\sqrt{13+4\sqrt{3}}=4+2\sqrt{3},$$ I.e.,
$$\sqrt{13+4\sqrt{3}}=1+2\sqrt{3},$$ which is
$$13+4\sqrt{3}=13+4\sqrt{3}.$$ Now just work backwards.
