Reformulate this task in the general case: for the positive distinct real numbers $a, b$ compare two numbers: $A = \sqrt{a+\sqrt{b}} + \sqrt{b+\sqrt{a}}$ and $B = \sqrt{a+\sqrt{a}} + \sqrt{b+\sqrt{b}}$. Prove that $A > B$.

Is equivalent to $A^2 > B^2 \Leftrightarrow \sqrt{a+\sqrt{b}} \cdot \sqrt{b+\sqrt{a}} > \sqrt{a+\sqrt{a}} \cdot \sqrt{b+\sqrt{b}}$.

After some simple calculations we have: $a\sqrt{a} + b\sqrt{b} > a\sqrt{b} + b\sqrt{a} \Leftrightarrow (a-b)(\sqrt{a}-\sqrt{b}) > 0$ if $a \ne b$.

This completes the proof.