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? 

$$f(x)=\sqrt{(6-x)(3+x)}+\sqrt{(3-x)(3+x)}+\sqrt{(3-x)^2}+\sqrt{3x(3-x)}$$
$\forall x\in [0,3]$, the factors $6-x$, $3+x$, $3-x$ and $3x$ in the radicands are non-negative.
By $GM\le AM$,
$$f(x)\le \frac{(6-x)+(3+x)}{2}+\frac{(3-x)+(3+x)}{2}+3-x+\frac{3x+(3-x)}{2}=12$$

Note that $12$ is not the global maximum, but a resonable upper bound.


