Prove if $\frac{a - b}{a + b} + \frac{b - c}{b + c} + \frac{c - a}{c + a} = 0$, then one of the terms is equal to $0$.

Question

The problem:

Numbers $a$, $b$, and $c$ are such, that $\frac{a - b}{a + b} + \frac{b - c}{b + c} + \frac{c - a}{c + a} = 0$. Prove that one of the terms is equal to $0$.

The textbook I'm using provides a hint, that unfortunately haven't brought me any closer to the solution yet:

Prove that $\frac{a - b}{a + b} + \frac{b - c}{b + c} + \frac{c - a}{c + a} = -\frac{(a - b)(b - c)(c - a)}{(a + b)(b + c)(c + a)}$

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Below are several outcomes that I was getting through the process of simplifying the initial equation, trying to achieve the result from the hint.

After expanding the equation:

$\frac{a}{{a + b}} - \frac{b}{{a + b}} + \frac{b}{{b + c}} - \frac{c}{{b + c}} + \frac{c}{{c + a}} - \frac{a}{{c + a}} \implies$

I'm resulting in this equation:

1. $\implies \frac{a(b - c)(b + c) \space - \space b(c - a)(c + a) \space - \space c(a - b)(a + b)}{(a + b)(c + a)(b + c)}$

or in this one:

2. $\implies \frac{ab(a + b) - bc(c - b) - ac(a + c)}{(a + b)(c + a)(b + c)}$

Another way I've tried is a direct simplification with a common denominator:

3. $\frac{{(a - b)(b + c)(c + a) + (b - c)(a + b)(c + a) + (c - a)(a + b)(b + c)}}{{(a + b)(b + c)(c + a)}}$

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Unfortunately, none of the approaches have led me to a conclusive proof.

I would greatly appreciate any guidance, insights, or alternative approaches that could help me prove the statement.

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*This problem is originally taken from a school's mathematics olympiad competition in my country and it has been challenging me for already a while.

Answer 1

Paraquoting,

Proposition: If $a,b,c\in\mathbb{R}\ $ and $\ a+b\neq 0;\ b+c\neq 0;\ c+a\neq 0,\quad $ then $$\quad \frac{a - b}{a + b} + \frac{b - c}{b + c} + \frac{c - a}{c + a} = -\frac{(a - b)(b - c)(c - a)}{(a + b)(b + c)(c + a)}.$$

Scrap work:

Well, if $\ a+b\neq 0;\ b+c\neq 0;\ c+a\neq 0,$ then both sides of the above equation exist. Moreover, we may multiply both sides of the equation by $(a + b)(b + c)(c + a)$ without losing solutions (we would have lost solutions if we multiplied by $0,$ but $(a + b)(b + c)(c + a)\neq 0$), to get:

$$ (a-b)(b+c)(c+a) + (b-c)(a+b)(c+a) + (c-a)(a+b)(b+c) = -(a - b)(b - c)(c - a). $$

Expanding the LHS and RHS give the same expression (I leave this as an exercise), and therefore "we're all good." However, this isn't a proof of the proposition, because we haven't started from known premises and arrived at the result of the proof as a conclusion. We have started assuming the proof and have arrived at an equation which we know is true. But now we have an idea of how to write a formal proof of the above proposition.

$$$$

Formal proof of above proposition:

Suppose $a,b,c\in\mathbb{R}\ $ and $\ a+b\neq 0;\ b+c\neq 0;\ c+a\neq 0.$

Expanding and simplifying the expression $(a-b)(b+c)(c+a) + (b-c)(a+b)(c+a) + (c-a)(a+b)(b+c)$ gives us [exercise to reader]. Expanding and simplifying the expression $-(a - b)(b - c)(c - a) $ gives us [exercise to reader]. We see that these two simplified expressions are equal/equivalent. It follows that

$$ (a-b)(b+c)(c+a) + (b-c)(a+b)(c+a) + (c-a)(a+b)(b+c) \equiv -(a - b)(b - c)(c - a). \quad (*)$$

Since $\ a+b\neq 0;\ b+c\neq 0;\ c+a\neq 0,$ we may divide both sides of $(*)$ through by $(a+b),$ then $(b+c),$ then $(c+a).$ This gives:

$$\frac{a - b}{a + b} + \frac{b - c}{b + c} + \frac{c - a}{c + a} \equiv -\frac{(a - b)(b - c)(c - a)}{(a + b)(b + c)(c + a)},$$

and the proposition has been proven.

Answer 2

The third expression, and considering the hint provided by @Adam Rubinson, looks like the best approach.

Considering the numerator, $$f=(a-b)(b+c)(c+a)+(b-c)(c+a)(a+b)+(c-a)(a+b)(b+c)$$

When you set $a=b$, i.e. replace all the $a$s with $b$s, you get$$0+(b-c)(c+b)(2b)+(c-b)(2b)(b+c)$$

Which is identically zero.

Therefore $(a-b)$ is a factor of $f$

By symmetry (i.e. by interchange of letters), then also $(b-c)$ and $(c-a)$ are also factors.

From this, and by inspection of coefficients, we see that $$f=(a-b)(b-c)(c-a)$$ and the rest of the question can now be completed.

Answer 3

Rewriting, $\frac{a - c}{c+ a} = \frac{a - b}{a + b} + \frac{b - c}{b + c} = \frac{ab - b^2 + ac - bc + ab - bc + b^2 -ac}{(a+b)(b+c)} = \frac{2b(a-c)}{(a+b)(b+c)}$

Now $a = c$ is one possibility and we are done.

But if $a \neq c$, we cancel $a-c$. We get:

$2bc + 2ba = ab + bc + ac + b^2$

$\implies b^2 - b(a+c) + ac = 0$

$\implies b = \frac{(a+c) \pm \sqrt{(a+c)^2 - 4ac}}{2}$

$\implies b = \frac{(a+c) \pm (a-c)}{2}$

$\implies b = a $ or $b = c$.