# Specialization Technique [closed]

In Beckenbach and Bellman's book Introduction to Inequalities, in the process of proving the generic Arithmetic\Geometric inequality, they have arrived at the proof for the generic

$$\displaystyle \frac{a_1+a_2+ \cdot \cdot\cdot + a_n}{n} \ge \sqrt[n]{a_1a_2\cdot\cdot\cdot a_n}$$

The process involves both forward and backwards induction. Turning to the 'backward' induction, they say this.

‘We accomplish this by the important technique of specialization ‘(sic), which they demonstrate thus.

Set $$a_1 = b_1,\space a_2 = b_2, \space a_3 = b_3$$ and ask for the value $$a_4$$ that yields the equality

$$\displaystyle \frac{a_1+a_2+ a_3 + a_4}{4} = \frac{b_1+b_2+ b_3}{3}$$

Question What is meant by specialization, as this term is not explained.

This is the edited question, and the answers below answer this for those, like me, working our way into this wonderful discipline.

• Do you have a specific question? Aug 1 at 23:24
• @JohnDouma. Well, I thought I had asked a specific question. The author/s mention a technique that is not explained, and googling it does not explain it either. Those steeped in math might think this trivial, but those of us trying to learn this discipline sometimes like to get an overview. Aug 1 at 23:28
• Asking for an overview of a topic is a request, not a question. The problem with your request is that it requires clairvoyance on the part of the person fulfilling the request. Aug 1 at 23:31
• @JohnDouma Such comments in a textbook can be quite confusing for the reader, I don't think this is a bad question. Aug 1 at 23:38

Specialization just means to take a general theorem, and look at some specific case. That's it. There's nothing deep here. Specialization is the opposite of generalization.

• The quadratic formula $$ax^2+bx+c=0\implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ is a general theorem. To solve the equation $$x^2+bx+6=0$$, you specialize by setting $$a=1,c=6$$. This gives a pair of solutions as a function of $$b$$.

• Your author had proved the general fact that $$\frac{a_1+a_2+a_3+a_4}4\ge \sqrt[4]{a_1a_2a_3a_4}$$. In particular, it is true in the case where $$a_1=b_1,\quad a_2=b_2,\quad a_3=b_3,\quad a_4=\frac{b_1+b_2+b_3}{3}$$ They are specializing to the case described above.

Specializing happens all the time, there's literally nothing interesting to say about it. It's not a surprising proof technique that needs to be taught, its just something mathematicians do all the time.

• Thank you very much. Aug 7 at 7:38

The method of proof used by that book using $$2^n$$ is also described in Wikipedia

The following proof by cases relies directly on well-known rules of arithmetic but employs the rarely used technique of forward-backward-induction. It is essentially from Augustin Louis Cauchy and can be found in his Cours d'analyse.[3]

The Wikipedia article does not mention the term "specialization" at all, and that does not seem to be a common name or used other than by those authors. Since the book you're referring to dates from 1961, it is quite likely that that naming did not take hold.

• Thank you Suzu. Aug 1 at 23:36
• Had I been able to accept more than one answer, I would have as ALL the answers are enlightening. Aug 10 at 8:56
• @user1115542 Mike Earnest's answer is probably the correct one. Aug 10 at 9:40
• Thank you Suzu. Aug 10 at 18:08

There are a lot of situations where we use this technique, but I've never heard the name 'specialization'. I'll give a few examples. It consists in writing our algebraic expression in a way we want and then finding the constants that arises.

1. We'd like to write the equation in the product form: $$\frac 1a+\frac 1b=\frac 1{17} \\ ab-17a-17b=0 \\ a(b-17)-17b=0.$$ We see that we need $$b-17$$ instead of $$b$$ in $$-17b$$, so we write what we want and then we correct it: $$a(b-17)-17((b-17)+17)=0\\ a(b-17)-17(b-17)=17^2\\ (a-17)(b-17)=17^2,$$ which can be easily solved for $$a,b\in\Bbb Z$$ integers. Here we didn'n need any new coefficients, since we did everything in our memory.

2. We would like to solve the quadratic equation not using formulae $$\Delta=\ldots$$, $$x_\pm=\frac{-b\pm\sqrt{\Delta}}{2a}$$. Here's our equation: $$x^2+4x-5=0.$$ Since $$x$$ is in two places we would like to use the formula $$(x+a)^2=x^2+2xa+a^2$$. We compare two first summands in our equation and this formula to get $$4x=2xa$$, so $$a=2$$. Now our equation takes the form $$(x+2)^2-4-5=0 \\ (x+2)^2=9,$$ which can be easily solved.

1. If we need to integrate rational functions we often use the method of equating coefficients to partial fraction decomposition. The example from Wikipedia is the following. We'd like to integrate the function $$\frac{1}{x(x-1)(x-2)}$$ so we'd like to present it in simpler form. We write $$\frac{1}{x(x-1)(x-2)} = \frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-2}$$ when $$A,B,C$$ are unknown coeficients that we can easily find. In the article there are more examples of this method. I think this is the same idea as in specialization technique.
• Thank you so much. Aug 7 at 7:37
• Had I been able to accept more than one answer, I would have as ALL the answers are enlightening. Aug 10 at 8:55