<s>[INST] Under certain conditions $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a'}+\frac{1}{b'}+\frac{1}{c'}\Rightarrow \{a,b,c\}=\{a',b',c'\}$ 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.
 [/INST] 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:
$$\frac{1}{3} + \frac{1}{6} + \frac{1}{4} = \frac{1}{5} + \frac{1}{20} + \frac{1}{2} = \frac{3}{4}$$
</s>