<s>[INST] How does one prove such an equation? The problem occurred to me while I was trying to solve a problem in planimetry using analytic geometry.
for $b$ between $-\frac{1}2$ and $1$ :
$\sqrt{2+\sqrt{3-3b^2}+b} = \sqrt{2-2b}+ \sqrt{2-\sqrt{3-3b^2}+b}$
 [/INST] Hint:
Square
$$\sqrt{2+\sqrt{3-3b^2}+b}- \sqrt{2-\sqrt{3-3b^2}+b} = \sqrt{2-2b}.$$
This gives
$$2+\sqrt{3-3b^2}+b+2-\sqrt{3-3b^2}+b-2\sqrt{(2+\sqrt{3-3b^2}+b)(2-\sqrt{3-3b^2}+b)}= 2-2b,$$
$$4+2b-2\sqrt{(2+b)^2-(3-3b^2)}= 2-2b,$$
$$2+4b-2\sqrt{4b^2+4b+1}= 0.$$
Remains to discuss the domain and the signs.
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