Under certain conditions $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a'}+\frac{1}{b'}+\frac{1}{c'}\Rightarrow \{a,b,c\}=\{a',b',c'\}$

Question

Let $a,b,c,a',b',c'\in \mathbb{Z}_{\geq 1}$ be such that $$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}<1,\quad \frac{1}{a'}+\frac{1}{b'}+\frac{1}{c'}<1. $$ Suppose $$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a'}+\frac{1}{b'}+\frac{1}{c'}. $$ Is it true that $\{a,b,c\}=\{a',b',c'\}$? If it is not, it should be easy to give a counterexample, but I am not able to find one.

Answer 1

COUNTEREXAMPLE.

Take $\left\{a,b,c \right\}=\left\{4k,6k,12k\right\}$ and $\left\{a',b',c' \right\}=\left\{6k,6k,6k\right\}$. Then both sums add up to $\frac{1}{2k}$.

Since @HRSE asks, here is another counterexample, in the case $\left\{a,b,c \right\}=\left\{40k,20k,120k\right\}$$\left\{a',b',c' \right\}=\left\{21k,42k,84k\right\}$ both sums add up to $\frac{1}{12k}$.

Answer 2

It is not true. Take a=2,b=3,c=-2,a'= 3, b'=3, c'=-3.

Answer 3

Since you did not specify that $a\neq b\neq c$ and $a'\neq b'\neq c'$ an obvious counterexample is:

$1/4 + 1/4 +1/3= 5/6$

$1/6+1/3+1/3= 5/6$

I think the interesting question is what if you assume that $a\neq b\neq c$. Does there then exist $a',b',c'$ such that $1/a+1/b+1/c=1/a'+1/b'+1/c'$, $a'\neq b'\neq c'$ and $\{a,b,c\}\neq \{a',b',c'\}$?

Answer 4

For the more general case when $a\neq b\neq c$ and $a'\neq b'\neq c'$, the answer is also no.

In particular, let $\frac1a +\frac1b = \frac{1}{a'}$, and $\frac{1}{c} = \frac{1}{b'} +\frac{1}{c'}$. Then $a' = \frac{ab}{a + b}$, and $c = \frac{b' c' }{b'+ c'}$. We'd want these to be integers.

A simple way of making $\frac{ab}{a + b}$ an integer is by setting one to $a = k+1$ and $b = k (k + 1)$. In this way, for $k = 2$, we can have $\frac{1}{3} + \frac{1}{6} = \frac{1}{2}$; and for $k = 4$, $\frac{1}{5} + \frac{1}{20} = \frac{1}{4}$. We have a counter-example:

$$\frac{1}{3} + \frac{1}{6} + \frac{1}{4} = \frac{1}{5} + \frac{1}{20} + \frac{1}{2} = \frac{3}{4}$$