# On The Maximization of Prod_{cyc}|(a-b)|

This is my third time submitting an article to the University of Computer Science, Romania. This time, I wrote the article on something new. Although the inequalities are somewhat classical-looking, I have tried to explain the logic behind maximization of the product, for nonnegative reals $a,b,c.$ I have attached the full article with this post, for those who are interested in seeing the whole article.

1. Introduction

Most inequalities that we come across have a simple type of equality case, ie $a=b=c$ or $a=b,c=0.$ But when we think of maximizing the product $(a-b)(b-c)(c-a)$ or its magnitude (both mean the same), the equality case cannot be as simple as $(1,1,1)$ because any two equal values of $a$ and $b$ can change the value of this expression directly to zero. Of course, considering the same thing over nonnegative reals and whole of reals is much different. So these types of inequalities usually have different equality cases which are not so obvious to determine. However, we will demonstrate how most of these inequalities can be easily solved with the help of AM-GM Inequality. After that, we will try to determine some general forms of this maximum, for different fixed parameters.

1.1 Pre-requisites

Here we assume that the reader has a good idea of the different ways of applying the AM-GM inequality, and the Cauchy-Schwarz inequality. These are the most basic and classical inequalities having monstrous applications. Also some basic knowledge about the $pqr$ and $uvw$ methods might come in handy, so that the reader can relate between these stuff easily. Basic knowledge on calculus, derivatives and solving polynomial equations is also required.

2. The $|(a-b)(b-c)(c-a)|$ problem

We will discuss this method by considering a well-known problem of Tran Quoc Anh(forum name, Nguoivn) which runs as follows.

If $a,b,c\geq 0,$ then we have $\displaystyle (a+b+c)^3\geq 6\sqrt 3 |(a-b)(b-c)(c-a)|.$

2.1. Determining the Equality Case

Firstly, we have to order $a,b,c$ so as to remove the modulus sign. Also, note that in these inequalities, it’s important to figure out an equality case somehow. In most of the cases, the equality occurs when one of the variables is set to zero. This is due to the nature of $(c-a)(c-b)(b-a)$ (assuming $c\geq b\geq a.$) When $a$ is the minimum of the three numbers, we see that the expression itself is a decreasing function in terms of $a,$ if we keep $b,c$ fixed. So this expression will be able to reach its maximum for a given value of $b,c$ if and only if $a$ is as small as possible, which is zero itself. Plugging in $a=0$ and $c=bt$ leads us to the cubic equation

$(t+1)^3=6\sqrt 3 t(t-1).$

A factorisation of this cubic expression gives us

$\displaystyle \left(t-\sqrt 3 -2\right)^2(t+7-4\sqrt{3})=0;$

Which leads us to the positive double root $t=\sqrt 3 +2.$

2.2. Significance of the Double Root

The inequality that we get after putting $a=0$ under the assumption $c\geq b\geq a$ is

$\displaystyle (b+c)^3\geq 6\sqrt{3}bc(c-b).$

Note that in this inequality, when the equality occurs for some value of $\dfrac cb,$ the curve $(t+1)^3-6\sqrt 3 t(t-1)$ must be just touching the $t$ axis at that point, so as to satisfy the inequality. Another consequence of this concept is that the other root will be on the negative side of the $t$ axis. Now, we can differentiate this equation to get the double root easily.

$\displaystyle \left[(t+1)^3-6\sqrt 3 t(t-1)\right]'=3(t+1)^2-12\sqrt 3 t-1=3(t-2-\sqrt 3)(t+4-3\sqrt 3).$

Now, we can check that which one of these is our double root by just plugging in these values into the previous equation.

2.3. The Equality Case

If we see that $c=\left(\sqrt{3}+2\right)b$ satisfies the conditions of equality, then the actual equality case turns out to be

$\displaystyle \boxed{(a,b,c)=\left(0,1,(2+\sqrt 3)\right)\equiv \left(0,\sqrt 3 -1 , \sqrt 3 +1\right)}$

The writing as conjugate surds is for aesthetic satisfaction.

2.4. Determining the Coefficients for AM-GM

Once we have determined the equality case, the rest is somewhat easy. Assuming that $c\geq b\geq a$ helps us writing $\displaystyle |(a-b)(b-c)(c-a)|\leq bc(c-b);$

And $(a+b+c)\geq (b+c);$ So that it is sufficient to check that

$\displaystyle (b+c)^3\geq 6\sqrt 3 bc(c-b).$

For this, we need to apply AM-GM in such a way that the three terms we choose on the right hand side are balanced. Using the equality case, we see that $(\sqrt 3+1)b=(\sqrt 3-1)c=c-b=2,$

And therefore the terms have been determined.

$\displaystyle 2bc(c-b)=\left[\left(\sqrt 3+1\right)b\right]\left[\left(\sqrt3-1\right)c\right]\left[(c-b)\right]\leq \frac{\left\{\sqrt 3(b+c)\right\}^3}{27}=\frac{(b+c)^3}{3\sqrt 3}.$

Since we have determined the equality case, therefore we need not worry about the right hand side. This method works mostly without fail, and is easy to implement. So, the method has worked out for $6\sqrt 3.$ We will discuss some more applications of this method. However, note that higher the degree of the inequality, the more unusable this method becomes.

3. Determination of Best Constants

In cases, these $|(a-b)(b-c)(c-a)|$ problems come with some determination of the “best constant” phrases. What it actually means is, say, we were talking of the inequality

$\displaystyle (a+b+c)^3\geq k|(a-b)(b-c)(c-a)|.$

Where $k$ is some positive real number, and $a,b,c\geq 0.$ In such a problem, they want us to figure out the largest value that $k$ can attain, because the inequality implies $k\leq\dfrac{(a+b+c)^3}{|(a-b)(b-c)(c-a)|}.$ If this is satisfied for all $a,b,c\geq 0,$ then we might try to minimize the right side first by minimizing $a,$ ie putting $a=0$ and $c=tb$ where $t\geq 1.$ Then this rewrites into

$\displaystyle k\leq \frac{(t+1)^3}{t(t-1)}.$

This is a single-variable inequality, so we may just want to differentiate this to find the minimum of $f(t)=\dfrac{(t+1)^3}{t(t+1)}.$ Simple calculus gives us the value $k=6\sqrt 3$ for $t=2+\sqrt 3.$ This way can also help us determine the equality case as we saw in the previous section. For some applications, we can get our hands on some problems. Note that this is, in cases, not complete enough to prove the original problem. There may be some other paramatrization in the problem, for example, an extra $(ab+bc+ca)$ term may be there, thus affecting the nature of the $|(a-b)(b-c)(c-a)|$ term, which was always decreasing in $\min\{a,b,c\}.$ But, this can always fetch us the best constants, and the equality cases in a packed fashion, so that the application of AM-GM becomes easier.

Caution: In an exam, we never mention the differentiations etc, just scribble the AM-GM step. All of this should be rough work.

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## 5 thoughts on “On The Maximization of Prod_{cyc}|(a-b)|”

1. That was delicious. Next article should be about hexagons because fuck yeah.

2. Potla says:

Well, next post will be on complete quadrilaterals, and next article will come up next year. 😛

• Potla says:

LOL, True… Wasn’t quite in the mood to change all that TeX formatting to WordPress-compatible stuff. 😀

• I’m curious what your definition of tl;dr is then.