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Question

Show that if $a, b, c$ are pairwise distinct real numbers and $$a ^ 3 b + b ^ 3 c + c ^ 3 a = a b ^ 3 + b c ^ 3 + c a ^ 3$$ then at least one of them is positive and at least one of them is negative.

My idea

First of all I put all the elements of the left member into the right member with different sign and i gave common factor so I obtained that:

$$a^3(b-c) + b^3(c-a) + c^3(a-b)=0.$$

We suppose that $a,b,c$ are all positive, and if we can show that this assumption is fals, we will know that at least one of them is negative

WLOG, $c < b < a$

$$a^3(b-c) + b^3(c-a) + c^3(a-b)=0$$

Because $b^3(c-a)$ is negative and the others are positive and summed they equal $0$, we have

$$a^3(b-c) + c^3(a-b)= -b^3(c-a)= b^3(a-c)=b^3(b-c)+b^3(a-b)$$ and because $c \leq b \leq a$ they cant be equal so this assumption is false, we know that at least one of them is negative.

Now, we suppose that $a,b,c$ are all negative, and if we can that this assumption is false, we will know that at least one of them is positive

WLOG, $a<b<c$

$a^3(b-c) + b^3(c-a) + c^3(a-b)=0$

Because $b^3(c-a)$ is positive and the others are negative we obteined that:

$a^3(b-c) + c^3(a-b)= -b^3(c-a)=b^3(a-c)$

From here is analogous from the first assumption.

I am not sure my idea is good. Hope one of you can tell me if my reasoning is good! Thank you!

Benjamin Wang
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IONELA BUCIU
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    1/ For solution-verification, you need to specify which step is in doubt. Please do so. – Calvin Lin Jan 10 '24 at 15:17
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    2/ For a different solution approach, try to factorize the expression. You came close to doing so when working with it, but never quite got to it. The idea is to show that $ a+b+c = 0 $, from which the conclusion is obvious. – Calvin Lin Jan 10 '24 at 15:21
  • @CalvinLin im not sure if the way i demonstrated that those assumptions are wrong is corect. – IONELA BUCIU Jan 11 '24 at 13:42
  • @CalvinLin i treid factorising and i got that $(b-c)(a-b)(a^2+ab-ac-c^2)=0$ because these are different numbers b-c and a-b cant be zero so the third member must equal 0. Idk what to do forward. – IONELA BUCIU Jan 11 '24 at 13:50
  • To be clear, I'm referring to factorizing the very first expression, and not what you've been working on. IE Factorize $ a^3b+b^3c+c^3a - ab^3 - bc^3 - ca^3$. Your factorization has the term $-a^2bc$, so it's not what I'm thinking of. – Calvin Lin Jan 11 '24 at 19:41
  • Hmm So you have $a^3k + c^3j = b^3k + b^3j$ where $k=(b-c);j=(a-b)$. Have to explain why that seems impossible. You have $a^3 > b^3$ but $b^3 > c^3$ so there can be some sort of averaging going on. – fleablood Jan 16 '24 at 17:40
  • Also you need to rule out what happens if one of them is $0$. – fleablood Jan 16 '24 at 17:43
  • If $a, b, c, < 0$ we can see that equality holds. Also, it is already true for $a, b, c, >0$. Furthermore, we know it is true for $a, b, c = 0$. I can't parse the question. What do you mean by at least one has to be positive and one has to be negative? – Agent Smith Jan 17 '24 at 04:26

3 Answers3

1

I'm going to first have a look on your idea, and then show another solution.

You are saying that if $0\lt c < b < a$, then $a^3(b-c) + c^3(a-b)$ and $b^3(b-c)+b^3(a-b)$ cannot be equal.

It seems to me that you are comparing $a^3(b-c)$ with $b^3(b-c)$, and $c^3(a-b)$ with $b^3(a-b)$.

We have $a^3(b-c)\gt b^3(b-c)$, but note that we have $c^3(a-b)\color{red}{\lt}b^3(a-b)$.

So, I don't know why you can say they cannot be equal. I think you need to add more explanations.

Another solution

$$a ^ 3 b + b ^ 3 c + c ^ 3 a = a b ^ 3 + b c ^ 3 + c a ^ 3$$

can be written as $$(b-c)a^3-(b-c)(b^2+bc+c^2)a+bc(b-c)(b+c)=0$$ So, we have $$(b-c)\bigg(a^3-(b^2+bc+c^2)a+bc(b+c)\bigg)=0$$ This can be written as $$(b-c)\bigg((c-a)b^2+c(c-a)b+a(a-c)(a+c)\bigg)=0$$ So, we have $$(b-c)(c-a)\bigg(b^2+cb-a(a+c)\bigg)=0$$ This can be written as $$(b-c)(c-a)\bigg((b-a)c+(b-a)(b+a)\bigg)=0$$ So, we finally have $$(b-c)(c-a)(b-a)(a+b+c)=0$$ from which the claim follows.

mathlove
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  • Thank you so much for your answer! May i ask you why from knowing that $a+b+c=0$ we get that at least one of them is positive and at least one of them is negative. – IONELA BUCIU Jan 17 '24 at 12:39
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    @IONELA BUCIU : If $a+b\gt 0$, then one of $a,b$ is positive, and $c=-a-b$ is negative. If $a+b=0$, then one of $a,b$ is positive, and the other is negative. If $a+b\lt 0$, then one of $a,b$ is negative, and $c=-a-b$ is positive. – mathlove Jan 17 '24 at 13:25
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Given that $a^3 b + b^3 c + c^3 a = a b^3 + b c^3 + c a^3$ with distinct $a,b,c\in\mathbb R$.

Consider a cubic polynomial

$$\begin{align*} P(x) &= a^3 b + b^3 x + x^3 a - a b^3 - b x^3 - x a^3\\ &= (a-b)x^3 + \left(b^3-a^3\right)x + a^3b - ab^3 \end{align*}$$

where $a$ and $b$ are the given constants, and $a-b\ne 0$.

Verify that:

  • $x=a$ is a root: $P(a) = a^3 b + b^3 a + a^3 a - a b^3 - b a^3 - a a^3 = 0$;
  • $x=b$ is a root, similarly; and
  • $x=c$ is a root from the given.

By Vieta's formula, the sum of roots of $P(x)$ is related to the coefficient of $x^2$:

$$a+b+c = 0$$

(This is the same result as in @mathlove's answer)

At least one of the pairwise sums of $a,b,c$ is non-zero, and the left out number has the opposite sign of the sum. (For example $a+b\ne 0$, then $c = -(a+b)$.)

peterwhy
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Let us assume $0 < a < b < c \implies 0 < a^3 < b^3 < c^3$

$a^3b + b^3c + c^3a = ab^3 + bc^3 + ca^3$

$a^3b < ca^3$ AND $c^3a < bc^3$

$a^3b + c^3a < bc^3 + bc^3$

Must be that, $b^3c > bc^3 \implies bc(b^2 - c^2) > 0 \implies bc(b - c)(b + c) > 0$.

That means $b > c$, which contradicts our assumption $0 < a < b < c$

So, $\neg (0 < a < b < c)$. They can't all be positive.

Let $a < b < c < 0$

$a^3b + b^3c + c^3a = ab^3 + bc^3 + ca^3$

$a^3b > ca^3$ AND $c^3a > bc^3$

$a^3b + c^3a > bc^3 + ca^3$

So, $b^3c < bc^3 \implies bc(b^2 - c^2) < 0$

That means $bc(b + c)(b - c) < 0$

$b < c \implies b - c < 0$

$0 < a < b < c \implies b + c < 0$

So, $bc < 0$ i.e. $(b < 0) \wedge (c > 0) \vee (b > 0) \wedge (c < 0)$. A contradiction when conjoined with our assumption $a < b < c < 0$

Ergo, not the case that for $a, b, c$, all are positive or all are negative.

Conclusion: Of $a, b, c$, some are positive and others are negative. QED

Agent Smith
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  • Same can be said for $a<0<c<b$, then $a^3b<ca^3$ AND $b^3c>ab^3$ AND $c^3a<bc^3$. Yet there are satisfying $(a,b,c)$, e.g. $a=-3, b=2,c=1$, and $b^3=8$ is a factor of $\left(ca^3+bc^3\right)-\left(a^3b+c^3a\right) = 32$. Then why does $a<0<c<b$ not lead to contradiction, but your $0<a<b<c$ case would? – peterwhy Jan 19 '24 at 18:51
  • Applying inequalities 1. $a^3b<ca^3$ and 2. $c^3a<bc^3$, what remains in $$\overbrace{a^3 b}^1 + b^3 c + \underbrace{c^3 a}_2 = a b^3 + \underbrace{b c^3}_2 + \overbrace{c a^3}^1$$ should be that $b^3c>ab^3$ -- not a contradiction with $0<a<b<c$, and same as in earlier revision. – peterwhy Jan 20 '24 at 22:50