# An Algebra question!

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• Jun 7th 2013, 09:28 PM
sam300
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,
Attachment 28547

then prove that,

Attachment 28548

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.
• Jun 7th 2013, 10:27 PM
chiro
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.
• Jun 7th 2013, 10:33 PM
Gusbob
Re: An Algebra question!
Quote:

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,
Attachment 28547

then prove that,

Attachment 28548

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 $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$. Your equation becomes $a^2+b^2+c^2=a+b+c$ and the question reduces to proving $a=b=c=1$

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

Some hints: Show $(a-b)^2+(b-c)^2+(a-c)^2=0$. Use the fact that $ab+bc+ac=a+b+c$.
• Jun 8th 2013, 05:58 AM
sam300
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,
Attachment 28550

OR by using (a+b+c)^2 identity:
Attachment 28551

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.
• Jun 8th 2013, 06:48 AM
Gusbob
Re: An Algebra question!
Quote:

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,
Attachment 28550

OR by using (a+b+c)^2 identity:
Attachment 28551

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: $(a-b)^2=a^2-2ab+b^2$ and similarly for $(b-c)^2$ and $(c-a)^2$. Adding the three together, one gets

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

Since $ab+bc+ca=a+b+c=a^2+b^2+c^2$, the right hand side is 0. As squares of real numbers are non-negative, it must be that $a-b=b-c=c-a=0$. That is, $a=b=c$. Finally, the equation $a^2+b^2+c^2=a+b+c$ means $3a^2=a^2$, so $a=0$ or $a=1$. We know the former case is impossible as it requires $x=0$, so we conclude that $a=b=c=1$
• Jun 8th 2013, 08:05 AM
sam300
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?
• Jun 8th 2013, 09:04 AM
Gusbob
Re: An Algebra question!
Quote:

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.
• Jun 9th 2013, 08:44 PM
sam300
Re: An Algebra question!
Bump, no one can do this?
• Jun 9th 2013, 09:38 PM
ibdutt
Re: An Algebra question!
• Jun 10th 2013, 02:08 AM
sam300
Re: An Algebra question!
Quote:

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.
• Jun 10th 2013, 07:19 PM
ibdutt
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.

Quote:

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.

• Jun 10th 2013, 08:14 PM
Idea
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)
• Jun 11th 2013, 07:38 AM
sam300
Re: An Algebra question!
Quote:

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?
• Jun 11th 2013, 08:43 AM
sam300
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.
• Jun 11th 2013, 08:43 PM
Idea
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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