# Using the hypothesis $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ to prove something else

Assuming that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$$

Is it possible to use this fact to prove something like:

$$\frac{1}{a^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=\frac{1}{a^{2013}+b^{2013}+c^{2013}}$$

Just curious. Thanks for the help!

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Well, assuming that $a,b,c\neq 0$, we can see that:

\begin{align} \frac{1}{a}+\frac{1}{b}+\frac{1}{c}&=\frac{1}{a+b+c} \\ \\ \frac{ab+bc+ac}{abc}&=\frac{1}{a+b+c} \\ \\ a^2b+abc+a^2c+ab^2+b^2c+abc+abc+bc^2+ac^2&=abc \\ \\ (ab+b^2+ac+bc)c+(ab+b^2+ac+bc)a&=0 \\ \\ (ab+b^2+ac+bc)(c+a)&=0 \\ \\ (a+b)(b+c)(c+a)&=0 \end{align}

Now we know that $(a+b)(b+c)(c+a)=0$

$\therefore a=-b\text{ or }b=-c\text{ or }a=-c$.

We can use the fact $a=-b$ to our LHS:

$$\frac{1}{(-b)^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=-\frac{1}{b^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=\frac{1}{c^{2013}}$$

$$\text{RHS: }\frac{1}{-b^{2013}+b^{2013}+c^{2013}}=\frac{1}{c^{2013}}=LHS$$

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+1. Though you only need to have one factor equal to $0$, right. Also, apart from $x\mapsto x^{2013}$, any function $x\mapsto f(x)$ with which is uneven, $f(-x)=-f(x)$, will give a relation. Or more generally, for fixed $b$ any function for which $f(-b)=-f(b)$. –  NikolajK May 17 '13 at 14:38
@DannyCheuk, change of side of one element from left to right will reduce calculation –  lab bhattacharjee May 17 '13 at 14:44
In the statement "$a=-b,b=-c,a=-c$" you need to say "or". In fact, just one or two of the conditions can be true, since $a$, $b$, and $c$ are nonzero. –  John Bentin May 17 '13 at 15:55
Thanks for pointing that out! –  user67258 May 17 '13 at 15:56
+1, though I would add the words "Without loss of generality" before $a = -b$. The reason is that it may well not be the case that $a = -b$, but we know at least one of $a = -b, b = -c$, or $a = -c$ occurs, and all cases are dealt with similarly. –  JavaMan May 17 '13 at 17:44

A simpler way to see that one of $a = -b$ or $b = -c$ or $c = -a$ is true.

wlog, we can assume $abc = 1$ (why?).

So we can assume $a,b,c$ are roots of $x^3 - px^2 + qx -1 = 0$.

Your relation gives us that $q = \frac{1}{p}$ (Vieta's formulas)

Thus we get $a,b,c$ are roots of

$$x^2(x - p) + \frac{1}{p}(x - p) = 0$$

Thus the roots are $$\pm\frac{1}{p}, p$$

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