# 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.

• Have you tried setting an optimisation problem and then solve it using the KKT condition? – RScrlli Apr 15 at 18:02

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

1. Two variables are equal.

Try to end this case by yourself.

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

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

• How can we prove it is less than or equal to 8? – RuaSun Apr 15 at 18:29
• I would use AM-GM. – Dr. Sonnhard Graubner Apr 15 at 18:47
• Can you elaborate how AM-GM helps to find the extrema of $\left( {a}^{2}+6\,ab+6\,ac+{b}^{2}+6\,bc+{c}^{2} \right) \left( {a}^{3}+3\,{a}^{2}b+3\,{a}^{2}c+3\,a{b}^{2}+2\,abc+3\,a{c}^{2} +{b}^{3}+3\,{b}^{2}c+3\,b{c}^{2}+{c}^{3} \right)$ under the given conditions $a+b+c=2$ and $c \ge 1$ ? – Martin R Apr 16 at 9:31