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 $a^2+b^2\leq 9, c=\sqrt{9-a^2}\sqrt{9-b^2}-2ab$
 A: 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 :
$\dfrac{9-a^2+9-b^2}{2} \ge \sqrt{(9-a^2)(9-b^2)} \implies 9- \dfrac{9}{2} \ge \sqrt{(9-a^2)(9-b^2)}$
$\dfrac{a^2+b^2}{2} \ge ab \implies 18 \ge 4ab$
A: $ 9a^2+9b^2+c^2=9a^2+9b^2+(9-a^2)(9-b^2)-4ab\sqrt{9-a^2}\sqrt{9-b^2}+4a^2b^2$
=$81+5a^2 b^2-4ab \sqrt{9^2-9(a^2+b^2)+a^2b^2} $
$\ge 81+5a^2 b^2-4|ab|\sqrt{9^2-9(a^2+b^2)+a^2 b^2} $     $  (-ab \ge -|ab|)$
$\ge 81+5a^2 b^2-4|ab|\sqrt{9^2-9(2|ab|)+a^2 b^2}$    $(a^2+b^2 \ge 2|ab|)$
=$81+5a^2 b^2-4|ab|(9-|ab|)$=$9a^2b^2-36|ab|+81$
$=9(|ab|-2)^2+45 \geq 45$ , $ (9(|ab|-2)^2 \ge 0)$
first "=", $ab \ge 0$, 2nd  "=", $|a|=|b|$, last "=", $|ab|=2$, so we got the min is 45 when $a=b=\pm \sqrt{2}$ 
with same method, we can get max also.
Edit: I add max in same way:
$81+5a^2 b^2-4ab \sqrt{9^2-9(a^2+b^2)+a^2b^2}  $
$\leq  81+5a^2 b^2+4|ab| \sqrt{9^2-9(a^2+b^2)+a^2b^2}$   $  (-ab \leq |ab|)$
$ \leq 81+5(\dfrac{a^2+ b^2}{2})^2+4*\dfrac{a^2+b^2}{2} \sqrt{9^2-9(a^2+b^2)+( \dfrac{a^2+b^2}{2} )^2} $  $ ( |ab| \leq \dfrac{a^2+b^2}{2}, a^2 b^2 \leq (\dfrac{a^2+b^2}{2})^2)$
$=81+5x^2+4x(9-x)=81+x^2+36x    $ ......... $  here:  x=\dfrac{a^2+b^2}{2} \leq \dfrac{9}{2}  $
$\leq 81+(\dfrac{9}{2})^2+36*\dfrac{9}{2}=\dfrac{567}{4}$
when $a=-b=\pm \dfrac{3\sqrt{2}}{2}$
A: Another approach - the symmetry here suggests Purkiss Principle (conditions to be verified), so the extremum is attained when $a = b$.  
So $c = (9-a^2) - 2a^2 = 9-3a^2$
and $9(a^2+b^2) + c^2 = 18a^2 + (9-3a^2)^2 = 9a^4 - 36a^2 + 81 = 9 (a^2 - 2)^2 + 45$  
which is minimised when $a^2 = 2$ or $a = \pm \sqrt2$.
