49th Mathematical Olympiad in Ukraine

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 $A^2>B^2\iff \sqrt{a+\sqrt{b}}\cdot \sqrt{b+\sqrt{a}}>\sqrt{a+\sqrt{a}}\cdot \sqrt{b+\sqrt{b}}$.

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

This completes the proof.