# A question involving inequality

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1. Jan 22, 2017

### Faiq

1. The problem statement, all variables and given/known data
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

3. 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.

2. Jan 22, 2017

### Staff: Mentor

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.

3. Jan 22, 2017

### Faiq

$\frac{\sum a^2(a-1)(b-1)(c-1)(d-1)}{(a-1)(b-1)(c-1)(d-1)(e-1)} = \frac{???}{1+p+q+r+s+t}$ 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: Jan 22, 2017
4. Jan 22, 2017

### haruspex

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.

5. Jan 22, 2017

### Ray Vickson

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: Jan 23, 2017