A question involving inequality

  • Thread starter Thread starter Faiq
  • Start date Start date
  • Tags Tags
    Inequality
AI Thread Summary
The discussion centers on proving the inequality a^2/(c-1) + b^2/(d-1) + c^2/(e-1) + d^2/(a-1) + e^2/(b-1) ≥ 20 for a, b, c, d, e > 1. Initial attempts involve polynomial equations and transformations, but participants express concerns about the validity of their methods, particularly regarding the direction of implications. Suggestions include leveraging symmetry and substituting variables to simplify the expressions. The concept of posynomials is introduced as a potential approach to find a minimum value for the function, indicating that optimization techniques may be necessary for a complete solution. The conversation emphasizes the complexity of the problem and the need for advanced mathematical tools.
Faiq
Messages
347
Reaction score
16

Homework Statement


If a,b,c,d,e>1
then prove that
a^2/(c-1)+b^2/(d-1)+c^2/(e-1)+d^2/(a-1)+e^2/(b-1)=>20

The Attempt at a Solution


Given a,b,c,d,e are roots of a polynomial equation of a degree 5 then
x^2/(x-1)+x^2/(x-1)+x^2/(x-1)+x^2/(x-1)+x^2/(x-1)=>20
5 x^2/(x-1)=>20
x^2/(x-1)=>4
x^2=>4x-4
x=> 2
This proves that a,b,c,d,e >1

I am sure this method is wrong because I prove Q implies P rather than proving P implies Q. However I cannot work out any other method.
 
Physics news on Phys.org
Faiq said:

Homework Statement


If a,b,c,d,e>1
then prove that
a^2/(c-1)+b^2/(d-1)+c^2/(e-1)+d^2/(a-1)+e^2/(b-1)=>20

The Attempt at a Solution


Given a,b,c,d,e are roots of a polynomial equation of a degree 5 then
x^2/(x-1)+x^2/(x-1)+x^2/(x-1)+x^2/(x-1)+x^2/(x-1)=>20
5 x^2/(x-1)=>20
x^2/(x-1)=>4
x^2=>4x-4
x=> 2
This proves that a,b,c,d,e >1

I am sure this method is wrong because I prove Q implies P rather than proving P implies Q. However I cannot work out any other method.
This, and the fact that you haven't shown that you may assume ##a=b=c=d=e=x##.
Not that I have an answer, but perhaps the symmetry (##f(a,b,c,d,e)=f(b,c,d,e,a)=\ldots##) of the quotients can be used somehow. I also think that the substitutions ##X := x-1## for ##x \in \{a,b,c,d,e\}## could help, e.g. to make a polynomial expression out of it.
 
##\frac{\sum a^2(a-1)(b-1)(c-1)(d-1)}{(a-1)(b-1)(c-1)(d-1)(e-1)} = \frac{?}{1+p+q+r+s+t}## where p,q,r,s,t are the coefficients of the equation having the roots a,b,c,d,e in the form ## x^5+px^4+qx^3+rx^2+sx+t##
Not sure how to find the question mark expression
 
Last edited:
I do not have a solution yet, but I suggest that substituting a+1 for a etc. everywhere would simplify the equations slightly.
Also, you might start with simpler cases, like 2 or 3 variables instead of 5. See if that provides any insight.

I got as far as needing to show that the only solution of the five simultaneous equations
2(a+1)a2=b(e+1)2
2(b+1)b2=c(a+1)2
etc. is a=b=c=d=e=1.
 
Faiq said:

Homework Statement


If a,b,c,d,e>1
then prove that
a^2/(c-1)+b^2/(d-1)+c^2/(e-1)+d^2/(a-1)+e^2/(b-1)=>20

The Attempt at a Solution


Given a,b,c,d,e are roots of a polynomial equation of a degree 5 then
x^2/(x-1)+x^2/(x-1)+x^2/(x-1)+x^2/(x-1)+x^2/(x-1)=>20
5 x^2/(x-1)=>20
x^2/(x-1)=>4
x^2=>4x-4
x=> 2
This proves that a,b,c,d,e >1

I am sure this method is wrong because I prove Q implies P rather than proving P implies Q. However I cannot work out any other method.

If we let ##a = 1 + x_1, b = 1 + x_2, \ldots, e = 1 + x_5##, we want ##x_i > 0## for ##i = 1, \ldots, 5##. We can write your function ##f = a^2/(c-1) + \cdots + e^2/(b-1)## as
$$f = (1+x_1)^2/x_3 + (1+x_2)^2/x_4 + (1+x_3)^2/x_5 + (1+x_4)^2/x_1 + (1+x_5)^2/x_2$$
We can expand out all the squares, collect terms, etc., to obtain a so-called posynomial in the variables ##x_1, x_2, x_3, x_4, x_5##. (A posynomial is a sum of terms of the form ##c\, x_1^{r_1} x_2^{r_2} x_3^{r_3} x_4^{r_4} x_5^{r_5}## with all coefficients ##c \geq 0##, but with positive, zero, or negative real numbers ##r_i## in the exponents, and of course with different ##\{r_i\}## and ##c## for different terms.) We want a minimum of ##f## over all the positive values of the ##x_i##. Although the function ##f## is, in general, a non-convex function of the ##x_j##, nevertheless, any stationary point of it is a global minimum! (The reason is that by substituting ##x_i = e^{y_i}## for all ##i## we get a function of ##(y_1, y_2, y_3, y_4, y_5)## that is strictly convex, so a stationary point is a global minimum.)

So, we can cheat a bit and use a computer optimization package to minimize ##f(y)## without worrying about issues of local/global minima. Or, we can write down the derivative conditions for a stationary point, and stare at those equations to see if a solution is "obvious". Or, we can just guess the form of solution and then verify that it "works", so is automatically the true solution.

In that way we can find the minimum possible value of ##f##---call it ##f_{\min}##-- and so know that ##f(a,b,c,d,e) \geq f_{\min}## for all allowed values of ##a\: \cdots\: e##.

I won't complete the analysis here because I don't want to spoil all your fun. Anyway, the solution proposed herein is definitely not in the "pre-calculus" category, and it even involves some concepts and facts that go beyond a first course in calculus.

BTW: concepts related to "posynomial" and the like first arose in so-called Geometric Programming, invented by Duffin, Peterson and Zener (the same Zener who invented the Zener diode). Some of the results and methods are very slick and sometimes even allow some complicated nonlinear optimization problems to be solved without using calculus at all---just algebra.
 
Last edited:
Back
Top