Can a^3+b^3+c^3 Equal d^3+e^3 Given Specific Conditions?

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Discussion Overview

The discussion revolves around the problem of comparing the sum of cubes, \(a^3+b^3+c^3\), with \(d^3+e^3\) under the constraints that the sums of squares and fourth powers of the variables are equal. Participants explore various approaches to tackle this problem, including algebraic manipulations and domain restrictions.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses frustration in solving the problem and seeks help with algebraic techniques.
  • Another participant suggests considering the limits of the variables, proposing scenarios where all variables are greater than 1, equal to 1, or between 0 and 1.
  • A participant derives a condition involving \(d\) based on eliminating \(e\) from the equations, leading to a quadratic inequality that must hold for \(d\) to be real.
  • There is a discussion about ruling out the case where all variables equal 1, as it leads to contradictions in the equations.
  • One participant proposes using trigonometric identities, such as letting \(a = \cos A\), to explore the scenario where all variables are between 0 and 1.
  • Participants note the implications of the derived inequalities but express confusion about the next steps in the problem-solving process.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the solution or the next steps. There are multiple competing views on how to approach the problem, and uncertainty remains regarding the implications of the derived conditions.

Contextual Notes

The discussion includes limitations related to assumptions about the variables and the implications of the derived inequalities, which are not fully resolved.

anemone
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Problem:

Let $a$, $b$, $c$, $d$ and $e$ be strictly positive real numbers such that:

$$a^2+b^2+c^2=d^2+e^2$$

$$a^4+b^4+c^4=d^4+e^4$$

Compare $$a^3+b^3+c^3$$ with $$d^3+e^3$$.

I have been exhausting all kinds of algebraic tricks, but I still don't get anywhere near to cracking it, and it is so frustrating not knowing how to solve it...

Any help and/or suggestions toward how to solve this problem is much appreciated.
 
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Have you considered limiting the Domain?

1) What happens when all of a, b, c, d, e are greater than 1?
2) Everything is 1
3) All are between 0 and 1.
 
First, I see some restrictions on $a, b$ and $c$. If you eliminate say $e$ from the pair of equations you get

$d^4 - (a^2+b^2+c^2)d^2 + a^2b^2 + a^2c^2 + b^2c^2 = 0$

Since $d$ is real this means that

$(a^2+b^2+c^2)^2-4(a^2b^2 + a^2c^2 + b^2c^2) \ge 0$

or

$a^4+b^4+c^4- 2 a^2b^2- 2 a^2c^2 -2 b^2c^2 \ge 0$

Also note that if $a=b=c = 1$ then we have a contradiction so this can't happen.
 
Last edited:
tkhunny said:
Have you considered limiting the Domain?

1) What happens when all of a, b, c, d, e are greater than 1?
2) Everything is 1
3) All are between 0 and 1.

Thanks for the hint, tkhunny...and I think we can rule out the possibility when all of them are 1 because

$$1^2+1^2+1^2 \ne 1^2+1^2$$ and $$1^4+1^4+1^4 \ne 1^4+1^4$$

But for the case where all of the variables are between 0 and 1, do you mean to suggest that we could let, for example a=cosA as another way to approach the problem?

Jester said:
First, I see some restrictions on $a, b$ and $c$. If you eliminate say $e$ from the pair of equations you get

$d^4 - (a^2+b^2+c^2)d^2 + a^2b^2 + a^2c^2 + b^2c^2 = 0$

Since $d$ is real this means that

$(a^2+b^2+c^2)^2-4(a^2b^2 + a^2c^2 + b^2c^2) \ge 0$

or

$a^4+b^4+c^4- 2 a^2b^2- 2 a^2c^2 -2 b^2c^2 \ge 0$

Also note that if $a=b=c = 1$ then we have a contradiction so this can't happen.

Thanks Jester for showing me some insight that I would not have thought of it myself...but, what should I do after that? I am so confused...
 

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