jmc

From the given equation, $$\sqrt{x+4\sqrt{x-4}}-\sqrt{x-4\sqrt{x-4}}=4.$$Squaring both sides, we get $$x+4\sqrt{x-4}-2\sqrt{x+4\sqrt{x-4}}\sqrt{x-4\sqrt{x-4}}+x-4\sqrt{x-4}=16.$$Hence, $$\begin{array}{rl}2x-16& =2\sqrt{(x+4\sqrt{x-4})(x-4\sqrt{x-4})}\\ & =2\sqrt{x^2-16(x-4)}\\ & =2\sqrt{x^2-16x+64}\\ & =2\sqrt{(x-8{)}^2}.\end{array}$$Equivalently, $x-8=\sqrt{(x-8{)}^2}.$ This holds if and only if $x\ge 8.$

All our steps are reversible, so the solution is $x\in \boxed{[8,\infty )}.$