Thread: An Algebra question!

1. An Algebra question!

Hey guys, I am quite new to this forum, and just encountered this seemingly simple problem but still was not successful in solving it.

If,

then prove that,

where x,y and z are real numbers.

I tried everything I could from basic identities to substituting x=a+y and x=b+z which one of my friend suggested but no luck yet.
The 'prove that' part seems obvious but I cannot make out how to write a proof.

I would appreciate if you guys can help me out with this.

2. Re: An Algebra question!

Hey sam300.

As a suggestion, you could re-write x,y,z as x* y* z* where x* = x + a, y* = y + b and z* = z + c and then simplify in terms of a, b, and c on one side. If you can prove that they have to be zero then you are done.

Also note that when you have squared terms, you know that things have to be positive.

A starting suggestion though would be firstly to get rid of the denominator on both sides by cross multiplying terms between the left and right hand sides.

In addition to this if you need more tools, take a look at the properties of norms if you have to. I don't think you will need to but norms are used with questions like this.

3. Re: An Algebra question!

Originally Posted by sam300
Hey guys, I am quite new to this forum, and just encountered this seemingly simple problem but still was not successful in solving it.

If,

then prove that,

where x,y and z are real numbers.

I tried everything I could from basic identities to substituting x=a+y and x=b+z which one of my friend suggested but no luck yet.
The 'prove that' part seems obvious but I cannot make out how to write a proof.

I would appreciate if you guys can help me out with this.
Let $\displaystyle a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$. Your equation becomes $\displaystyle a^2+b^2+c^2=a+b+c$ and the question reduces to proving $\displaystyle a=b=c=1$

This should be a much easier problem to work with. See if you make any headway here.

Some hints: Show $\displaystyle (a-b)^2+(b-c)^2+(a-c)^2=0$. Use the fact that $\displaystyle ab+bc+ac=a+b+c$.

4. Re: An Algebra question!

chiro,

I did not quite get you, are you saying I should get rid of denominators and substitute x as x'-a (or x'+a?)? How can I put x'=x+a?
I am sorry, I'm not acquainted with the properties of norms as you are saying.

Gusbob,

I tried your approach but no success yet,
at most I got,

OR by using (a+b+c)^2 identity:

But I'm stuck at that. Thanks for your replies anyway, please do reply if you have any idea what to do next.

PS. I have no idea why the latex is appearing so small.

5. Re: An Algebra question!

Originally Posted by sam300
chiro,

I did not quite get you, are you saying I should get rid of denominators and substitute x as x'-a (or x'+a?)? How can I put x'=x+a?
I am sorry, I'm not acquainted with the properties of norms as you are saying.

Gusbob,

I tried your approach but no success yet,
at most I got,

OR by using (a+b+c)^2 identity:

But I'm stuck at that. Thanks for your replies anyway, please do reply if you have any idea what to do next.

PS. I have no idea why the latex is appearing so small.
Quick sketch of proof: $\displaystyle (a-b)^2=a^2-2ab+b^2$ and similarly for $\displaystyle (b-c)^2$ and $\displaystyle (c-a)^2$. Adding the three together, one gets

$\displaystyle (a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2)-2(ab+bc+ca)$

Since $\displaystyle ab+bc+ca=a+b+c=a^2+b^2+c^2$, the right hand side is 0. As squares of real numbers are non-negative, it must be that $\displaystyle a-b=b-c=c-a=0$. That is, $\displaystyle a=b=c$. Finally, the equation $\displaystyle a^2+b^2+c^2=a+b+c$ means $\displaystyle 3a^2=a^2$, so $\displaystyle a=0$ or $\displaystyle a=1$. We know the former case is impossible as it requires $\displaystyle x=0$, so we conclude that $\displaystyle a=b=c=1$

6. Re: An Algebra question!

Well you are correct, and I had been trying to arrive at same proof.

But how did you assume that ab+bc+ca=a+b+c?

7. Re: An Algebra question!

Originally Posted by sam300
Well you are correct, and I had been trying to arrive at same proof.

But how did you assume that ab+bc+ca=a+b+c?
Sorry I made a mistake here. I will fix it tomorrow morning.

8. Re: An Algebra question!

Bump, no one can do this?

10. Re: An Algebra question!

Originally Posted by ibdutt
I have already tried that method, but to say that x=y=z we have to have squares on LHS.
Your answer also has cubes and expressions like (x-y) which may make the term negative
Thus when some terms are negative and some positive, their sum can be zero.

11. Re: An Algebra question!

the thinking is good. i would like to mention that on the LHS we have sum squares of rationals and so that will always be positive.
On the RHS we have the sum of the same rationals so the sign of x,y and z has to be the same.
I hope that is oK.

Originally Posted by sam300
I have already tried that method, but to say that x=y=z we have to have squares on LHS.
Your answer also has cubes and expressions like (x-y) which may make the term negative
Thus when some terms are negative and some positive, their sum can be zero.

12. Re: An Algebra question!

The stated problem is equivalent to a^2 + b^2 + c^2 = a + b + c and abc=1

We have (a-1)^2 + (b-1)^2 + (c-1)^2 + (a + b + c) = 3 which shows that a + b + c <= 3

If a,b,c are >0 then we also have a + b + c >= 3 (Arithmetic average is greater than the geometric average)

13. Re: An Algebra question!

Originally Posted by Idea
The stated problem is equivalent to a^2 + b^2 + c^2 = a + b + c and abc=1

We have (a-1)^2 + (b-1)^2 + (c-1)^2 + (a + b + c) = 3 which shows that a + b + c <= 3

If a,b,c are >0 then we also have a + b + c >= 3 (Arithmetic average is greater than the geometric average)
You are a genius dude, but there is one little important flaw.
How did you assume a,b,c>0, they can be -ve also, can't they?

14. Re: An Algebra question!

And due to your arithmetic mean and geometric mean hint,
I found a simpler way,
As, abc=1
Each of a,b,c is -1 or +1,
So we have three cases,
1) All three are +1
2) All three are -1
3) Any two are -1 and one is +1

Try each case in,
(a+b+c)/3>=(abc)^(1/3) OR (a+b+c)/3>= 1 (As abc=1)

And you will get the answer, just the flaw is we don't know if a,b,c>0.
And therefore this is not a perfect proof. But it is a proof nonetheless.

15. Re: An Algebra question!

abc=1 does not imply a,b,c is -1 or +1 as these are real numbers, e.g. a=1/3, b=2, c=3/2 , abc=1
As for the other question, a,b,c>0 is not an assumption. It follows from the given.

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