# 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? – Michael Rozenberg Aug 22 '20 at 10:21

## 4 Answers

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? – Book Of Flames Aug 22 '20 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? – Book Of Flames Aug 22 '20 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) – Book Of Flames Aug 22 '20 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. – Michael Rozenberg Aug 22 '20 at 10:28
• Vasc's solution is nice. (+1) – River Li Aug 22 '20 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: $$\prod_{cyc}(a^2-a+1)\geq\prod_{cyc}\sqrt{\frac{a^6+a^3+1}{3}}\geq\frac{1}{3}(a^2b^2c^2+abc+1).$$

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) – River Li Aug 22 '20 at 15:22
• How is Holder's Inequality applicable when a,b,c are all reals? I guess it works for positive reals only. – Book Of Flames Aug 22 '20 at 15:22
• @Book Of Flames I added something. We can work with other cases by the similar way. – Michael Rozenberg Aug 22 '20 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. – Michael Rozenberg Aug 22 '20 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. – River Li Aug 22 '20 at 15:26
• how to find this SOS? – tthnew Aug 22 '20 at 22:45
• For me, this inequality is hard to get SOS. – tthnew Aug 22 '20 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). – River Li Aug 23 '20 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, $$\begin{equation}\begin{split} 1+a^2b^2 \leq& 2AB \\ =& 2(1-a+a^2)(1-b+b^2), \\ \end{split}\end{equation}$$ which is also used in Vasc's solution provided by Michael Rozenberg.