Find minimum in a constrained two-variable inequation I would appreciate if somebody could help me with the following problem: 
Q: find minimum 
$$9a^2+9b^2+c^2$$
where $a^2+b^2\leq 9, c=\sqrt{9-a^2}\sqrt{9-b^2}-2ab$

Maybe this comes to your rescue. 
Consider $b \ge a \ge 0$ 
When you expansion of $(\sqrt{9-a^2}\sqrt{9-b^2}-2ab)^2=(9-a^2)(9-b^2)+4a^2b^2-4ab \sqrt{(9-a^2)(9-b^2)}$
This attains minimum when $4ab \sqrt{(9-a^2)(9-b^2)}$ is maximum. 
Applying AM-GM :
$\dfrac{9-a^2+9-b^2}{2} \ge \sqrt{(9-a^2)(9-b^2)} \implies 9- \dfrac{9}{2} \ge \sqrt{(9-a^2)(9-b^2)}$
$\dfrac{a^2+b^2}{2} \ge ab \implies 18 \ge 4ab$
