Elementary Algebra Problem (in 8th grade) The exercise is to prove that $$ \forall x \in [0,3] $$ : $$ f(x)=\sqrt{18 + 3x -x^{2}} + \sqrt{9-x^{2}} + \sqrt{9-6x+x^{2}} + \sqrt{9x-3x^{2}} \le 12 $$
I notice that when $$ x=0 \implies f(x) = 3\sqrt{2} + 3 + 3 \le 12 $$ 
and when $$ x=3 \implies f(x) = 3\sqrt{2} \le 12 $$, but how further? 

Hints: 
Each radicand is a quadratic polynomial which has a maximum (the corresponding parabola is oriented downwards) or a minimum  (the corresponding parabola is oriented upwards). The extremum of $ax^2+bx+c\;$ is obtained at  $x=-\dfrac b{2a}$.
