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Compare a³+b³+c³ with d³+e³

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
Problem:

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

\(\displaystyle a^2+b^2+c^2=d^2+e^2\)

\(\displaystyle a^4+b^4+c^4=d^4+e^4\)

Compare \(\displaystyle a^3+b^3+c^3\) with \(\displaystyle d^3+e^3\).

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.
 

tkhunny

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MHB Math Helper
Jan 27, 2012
267
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.
 

Jester

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MHB Math Helper
Jan 26, 2012
183
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.
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,756
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

\(\displaystyle 1^2+1^2+1^2 \ne 1^2+1^2\) and \(\displaystyle 1^4+1^4+1^4 \ne 1^4+1^4\)

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?

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...