Find the minimum and maximum values of $P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$ Let $a,b,c$ be non-negative real numbers such that $c \geq 1$ and that $a+b+c=2$. Find the minimum and maximum values of
$$P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$$
To find the minimum of $P$ I rewrite it as
$$P=2\left( ab+bc+ca+1\right)\left(2-abc\right)$$
Then I show that $P \geq 4$, equivalently show that
$$2\left( ab+bc+ca\right)-abc\left( ab+bc+ca\right) \geq abc$$
Indeed, we have
$$3abc\left( a+b+c\right) \leq \left( ab+bc+ca\right)^2 \Leftrightarrow 6abc \leq \left( ab+bc+ca\right)^2 $$
Thus we need to show
$$ \left( ab+bc+ca\right)^2 \leq 12\left( ab+bc+ca\right) -6abc\left( ab+bc+ca\right)$$
This is equivalent to
$$\left( ab+bc+ca\right)\left( ab+bc+ca+6abc-12\right) \leq 0$$
This implies from the inequalities that $ab+bc+ca \leq \left( a+b+c\right)^2/3=4/3$ and $abc \leq \left( a+b+c\right)^3/27=8/27$. The equality holds if $a=b=0$ and $c=1$ 
Is this right? And for the maximum value, I have no idea. I only guess it is $8$ attending at $a=0,b=c=1$. Please help me. Thank you.

I would use that $$6=\frac{3}{2}(a+b+c)^2$$ and $$2=\frac{1}{4}(a+b+c)^3$$
Using this substitution we get the term
$$1/8\, \left( {a}^{2}+6\,ab+6\,ac+{b}^{2}+6\,bc+{c}^{2} \right) 
 \left( {a}^{3}+3\,{a}^{2}b+3\,{a}^{2}c+3\,a{b}^{2}+2\,abc+3\,a{c}^{2}
+{b}^{3}+3\,{b}^{2}c+3\,b{c}^{2}+{c}^{3} \right) 
$$
