# Proving $3(1−a+a^2)(1−b+b^2)(1−c+c^2)≥1+abc+a^2b^2c^2$

My task was to prove the question above over real variables.

I thought that this minor inequality should help- $$3(1 − a + a^2)(1 − b + b^2) ≥ 2(1 − ab + a^2 b^2).$$ which is true.

By this inequality, the original inequality is converted to- $$(1 - ab)^2 (1 - c)^2 + (ab-c)^2 + abc \geq 0$$ This proves the inequality for $$abc\geq 0$$.
I want to prove this Inequality for $$abc\lt0$$. But I couldn't find a solution for $$abc\lt0$$.

Any extensions for $$abc\lt0$$ are thankfully accepted.

• Can down-voter explain us, why did you do it? Aug 22, 2020 at 10:21

Your first step leads to a wrong inequality because it does not save the case of the equality occurring: $$a=b=c=1.$$

After your first step it's enough to prove that: $$2(1-ab+a^2b^2)(1-c+c^2)\geq1+abc+a^2b^2c^2,$$ which is wrong for $$a=b=c=1.$$

The Vasc's solution.

Since $$2(a^2-a+1)(b^2-b+1)\geq a^2b^2+1,$$ it's enough to prove that: $$3(a^2b^2+1)(c^2-c+1)\geq2(a^2b^2c^2+abc+1),$$ which is a quadratic inequality of $$c$$.

Can you end it now?

• Well, it is a good solution, but it's difficult to see $2(a^2-a+1)(b^2-b+1)\geq a^2b^2+1$ could do the proof. Is there something more straightforward, something like the extension of my attempt? Aug 22, 2020 at 10:07
• Why does the solution go wrong if the equality case wasn't saved? After all it's just to be proved true, how does the equality case matters? Aug 22, 2020 at 10:14
• I think you got $2(a^2-a+1)(b^2-b+1)\geq a^2b^2+1$ from an identity which contains four terms, how did you see it before that leaving the 2 square terms does the job? (While leaving $a^2b^2+1$ as is) Aug 22, 2020 at 10:25
• @Book Of Flames It seems that $2(a^2-a+1)(b^2-b+1)-a^2b^2-1$ may be a sum of squares. I just tried to get it and it turned out. I did not see it before, of course. Aug 22, 2020 at 10:28
• Vasc's solution is nice. (+1) Aug 22, 2020 at 15:10

Another way.

It's enough to prove our inequality for non-negatives $$a$$, $$b$$ and $$c$$.

Now, since $$3(a^2-a+1)^3-a^6-a^3-1=(a-1)^4(2a^2-a+2)\geq0,$$ by Holder we obtain: $$\prod_{cyc}(a^2-a+1)\geq\prod_{cyc}\sqrt[3]{\frac{a^6+a^3+1}{3}}\geq\frac{1}{3}(a^2b^2c^2+abc+1).$$

Now, let $$a\leq0$$, $$b\geq0$$ and $$c\geq0.$$

Thus, after replacing $$a$$ on $$-a$$ we need to prove that: $$3\sum_{cyc}(a^2+a+1)(b^2-b+1)(c^2-c+1)\geq a^2b^2c^2-abc+1,$$ which follows from the previous inequality: $$3\sum_{cyc}(a^2+a+1)(b^2-b+1)(c^2-c+1)\geq$$ $$\geq3\sum_{cyc}(a^2-a+1)(b^2-b+1)(c^2-c+1)\geq a^2b^2c^2+abc+1\geq a^2b^2c^2-abc+1.$$

• It is very nice. (+1) Aug 22, 2020 at 15:22
• How is Holder's Inequality applicable when a,b,c are all reals? I guess it works for positive reals only. Aug 22, 2020 at 15:22
• @Book Of Flames I added something. We can work with other cases by the similar way. Aug 22, 2020 at 15:35

Two SOS solutions with the help of computer

1. According to Vasc's solution in @Michael Rozenberg's answer, we have a simple SOS expression: \begin{align} &3(a^2-a+1)(b^2-b+1)(c^2-c+1) - (1 + abc + a^2b^2c^2)\\ =\ & \frac{1}{8}(abc-3c+2)^2 + \frac{3}{8}(abc-2ab+c)^2 + \frac{3}{8}(a-1)^2(b-1)^2(2c-1)^2\\ &\quad + \frac{9}{8}(a-1)^2(b-1)^2 + \frac{3}{8}(a-b)^2(2c-1)^2 + \frac{9}{8}(a-b)^2. \end{align}

2. Without using Vasc's solution, I can obtain a complicated SOS expression $$3(a^2-a+1)(b^2-b+1)(c^2-c+1) - (1 + abc + a^2b^2c^2) = \frac{1}{2}z^\mathsf{T}Qz$$ where $$z = [1, a, b, c, ab, ca, bc, abc]^\mathsf{T}$$ and $$Q = \left(\begin{array}{rrrrrrrr} 4 & -3 & -3 & -3 & 2 & 2 & 2 & -1\\ -3 & 6 & 1 & 1 & -3 & -3 & -1 & 2\\ -3 & 1 & 6 & 1 & -3 & -1 & -3 & 2\\ -3 & 1 & 1 & 6 & -1 & -3 & -3 & 2\\ 2 & -3 & -3 & -1 & 6 & 1 & 1 & -3\\ 2 & -3 & -1 & -3 & 1 & 6 & 1 & -3\\ 2 & -1 & -3 & -3 & 1 & 1 & 6 & -3\\ -1 & 2 & 2 & 2 & -3 & -3 & -3 & 4 \end{array}\right).$$ Remarks: $$Q$$ is positive semidefinite.

• It's interesting! Does your method help for any sixth degree homogeneous polynomial of three variables? With real coefficients. Aug 22, 2020 at 15:19
• @MichaelRozenberg First, I can only find SOS case-by-case. Second, often I can not find simple SOS. Third, usually some solutions are simpler/better than SOS, e.g., your solution by Holder for this inequality. Aug 22, 2020 at 15:26
• how to find this SOS? Aug 22, 2020 at 22:45
• For me, this inequality is hard to get SOS. Aug 22, 2020 at 22:50
• @tthnew Using sos solvers (I use sostools in matlab, but I think that other solvers not in matlab also work), you can get $Q$ which is a matrix of real numbers (but in this question, $Q$ is almost rational). Aug 23, 2020 at 0:51

Here is a straightforward solution based on quadratic inequalities.

For simplicity, denote $$A=1-a+a^2$$ and $$B=1-b+b^2$$. We need to show that $$3AB(1-c+c^2) \geq 1+abc +a^2b^2c^2.$$ This is equivalent to showing that $$(3AB-a^2b^2)c^2 - (3AB+ab)c + 3(AB-1) \geq 0.$$ If we regard the left-hand side above as a quadratic function of $$c$$, $$f_{A,B,a,b}(c)= (3AB-a^2b^2)c^2 - (3AB+ab)c + 3(AB-1),$$ it suffices to show $$f_{A,B,a,b}(c)\geq 0$$ for any real $$a,b,c$$. From the fact $$A=1-a+a^2 \geq \frac 3 4 a^2$$ and $$B\geq \frac 3 4 b^2$$, we know the leading coefficient of $$f_{A,B,a,b}$$ is strictly positive, i.e., $$3AB - a^2b^2 >0$$. Now it remains to show the discriminant of $$f_{A,B,a,b}$$ is non-positive. Namely, $$(3AB +ab)^2 -4(3AB-a^2b^2)(3AB-1) \leq 0,$$ or equivalently, $$4AB + 2ABab + 4ABa^2b^2 \leq a^2b^2 +9A^2B^2.$$ By AM-GM inequality, we have $$2ABab \leq a^2b^2 + A^2B^2.$$ Therefore, it suffices to show $$4AB + 4ABa^2b^2 \leq 8A^2B^2,$$ or equivalently, $$$$\begin{split} 1+a^2b^2 \leq& 2AB \\ =& 2(1-a+a^2)(1-b+b^2), \\ \end{split}$$$$ which is also used in Vasc's solution provided by Michael Rozenberg.