Austria 2010

Since we are given that $0\le a,b\le 1$ holds, it follows from the AM-GM inequality, that $$\begin{array}{rl}& \sqrt{a^3{b}^3}+\sqrt{(1-a^2)(1-ab)(1-b^2)}\\ & \le \sqrt[3]{a^3{b}^3}+\sqrt[3]{(1-a^2)(1-ab)(1-b^2)}\\ & =\sqrt[3]{a^2\cdot ab\cdot b^2}+\sqrt[3]{(1-a^2)(1-ab)(1-b^2)}\\ & \le \frac{a^2+ab+b^2}{3}+\frac{(1-a^2)+(1-ab)+(1-b^2)}{3}\\ & =1,\end{array}$$ as claimed. We note that equality holds iff $a^2=ab=b^2$ and either $a^3{b}^3=0$ or $a^3{b}^3=1$ holds (and therefore also the same for $(1-a^2)(1-ab)(1-b^2)$). Equality therefore holds iff $a=b=0$ or $a=b=1$.