<s>[INST] 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.
 [/INST] $$\sqrt{3+\sqrt{13+4\sqrt{3}}}= \sqrt{3+\sqrt{{(2\sqrt{3}+1)}^2}} = \sqrt{4+ 2\sqrt{3}} = \sqrt{{(\sqrt{3}+1)}^2} = \sqrt{3}+1.$$
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