<s>[INST] 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.
 [/INST] Let $a=b=0$ and $c=2$. 
Thus, $P=4$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$(6-a^2-b^2-c^2)(2-abc)\geq4$$ or
$$\left(\frac{3(a+b+c)^2}{2}-a^2-b^2-c^2\right)\left(\frac{(a+b+c)^3}{4}-abc\right)\geq\frac{(a+b+c)^5}{8}$$ or
$$\sum_{cyc}(a^4b+a^4c+3a^3b^2+3a^3c^2+6a^3bc+6a^2b^2c)\geq0,$$ which is obvious.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$ and $M$ is a maximal value.
Thus, the inequality $P\leq M$ is a linear inequality of $w^3$, which says that it's enough to prove this for the extreme value of $w^3$, which happens in the following cases.


*

*$c=1$.


In this case $b=1-a$, where $0\leq a\leq1$ and $$P=8-2(a-a^2)^2\leq8;$$
2. $w^3=0$.
Let $b=0$, $c=2-a,$ where $2-a\geq1,$ which says $0\leq a\leq1.$
Thus, $$P=8-4(a-1)^2\leq8.$$


*Two variables are equal. 


Try to end this case by yourself. 
Actually, this case we can end by AM-GM. 
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