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The expression $$\frac{2(\sqrt{2} + \sqrt{6})}{3\sqrt{2 + \sqrt{3}}}$$ is equal to
$$\frac{2(\sqrt{2} + \sqrt{6})}{3\sqrt{2 + \sqrt{3}}}$$
Let's rationalize the denominator.
$$\frac{2(\sqrt{2}+\sqrt{6})}{3\sqrt{2+\sqrt{3}}}\cdot\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2-\sqrt{3}}}$$