I recognize the expression on the left as a nested radical, and I wonder if I can simplify it by finding a perfect square inside the inner square root. I notice that $21 + 12 \sqrt{3}$ is the same as $(3 + 2 \sqrt{3})^2,$ since expanding the latter gives $9 + 12 \sqrt{3} + 12 = 21 + 12 \sqrt{3}.$ So I can rewrite the expression as $\sqrt{1 + \sqrt{(3 + 2 \sqrt{3})^2}}.$ Now I can use the property that $\sqrt{a^2} = |a|,$ and since $3 + 2 \sqrt{3}$ is positive, I can take off the absolute value bars. This gives me $\sqrt{1 + 3 + 2 \sqrt{3}} = \sqrt{4 + 2 \sqrt{3}}.$ I still want to simplify this further, and I notice that $4 + 2 \sqrt{3}$ is the same as $(1 + \sqrt{3})^2,$ since expanding the latter gives $1 + 2 \sqrt{3} + 3 = 4 + 2 \sqrt{3}.$ So I can rewrite the expression again as $\sqrt{(1 + \sqrt{3})^2}.$ And again, using the property that $\sqrt{a^2} = |a|,$ and since $1 + \sqrt{3}$ is positive, I can take off the absolute value bars. This gives me $1 + \sqrt{3}.$ So I have simplified the left-hand side of the equation to $1 + \sqrt{3},$ and I want to find the ordered pair $(a,b)$ of positive integers such that $\sqrt{a} + \sqrt{b} = 1 + \sqrt{3}.$ If that is the case, then I need to find $a$ such that $\sqrt{a} = 1.$ The only positive integer that satisfies this is $a = 1.$ So I have found the ordered pair $(a,b)$ of positive integers that satisfies the equation. 
Find the ordered pair $(a,b)$ of positive integers, with $a < b,$ for which
\[\sqrt{1 + \sqrt{21 + 12 \sqrt{3}}} = \sqrt{a} + \sqrt{b}.\]