Find the minimum and maximum values of $P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$

Question

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.

Answer 1

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 $$1/8\, \left( {a}^{2}+6\,ab+6\,ac+{b}^{2}+6\,bc+{c}^{2} \right) \left( {a}^{3}+3\,{a}^{2}b+3\,{a}^{2}c+3\,a{b}^{2}+2\,abc+3\,a{c}^{2} +{b}^{3}+3\,{b}^{2}c+3\,b{c}^{2}+{c}^{3} \right) $$

Answer 2

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.

1. $c=1$.

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, $$P=8-4(a-1)^2\leq8.$$

3. Two variables are equal.

Try to end this case by yourself.

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

Answer 3

Let $q = ab$.

We have \begin{align*} P &= (6 - (2-c)^2 + 2q - c^2)(2 - qc)\\ &= -2cq^2 - 2(c-1)(2-c)(c+1)q - 4(c-1)^2 + 8\\ &\le 8 \end{align*} with equality if $c = 1$ and $q = 0$.

On the other hand, we have \begin{align*} P &= (6 - (2-c)^2 + 2q - c^2)(2 - qc)\\ &= -2cq^2 - 2(c-1)(2-c)(c+1)q - 4(c-1)^2 + 8\\ &\ge -2c \left(\frac{(2 - c)^2}{4}\right)^2 - 2(c-1)(2-c)(c+1)\cdot \frac{(2 - c)^2}{4} - 4(c-1)^2 + 8\\ &= \frac38\,{c}^{5}-2\,{c}^{4}+\frac52\,{c}^{3}-{c}^{2}+8\\ &\ge 4 \end{align*} with equality if $c = 2$ and $q = 0$, where we use $q \le (a + b)^2/4 = (2-c)^2/4$ and $$\frac38\,{c}^{5}-2\,{c}^{4}+\frac52\,{c}^{3}-{c}^{2}+8 - 4 = \frac18(2-c)(-3c^4 + 10c^3 + 8c + 16) \ge 0.$$

We are done.