Question:
What are all the solutions to $$a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ for fractions/rational numbers $a,b,c\in\mathbb{Q}$?
I asked a similar question a few weeks ago with perfect squares with no success. So I thought to try an easier version of the problem by ditching the perfect squares requirement.
Known solutions:
If any of the three rational numbers is $\pm1$, then our solution is part a simple infinite family of solutions, $$(a,b,c)\quad=\quad(\pm1,t,1/t)\quad\text{or}\quad(\pm1,t,-t)$$ for some rational $t\in\mathbb{Q}$.
But finding solutions where $a,b,c\not=\pm 1$ seems harder. By brute force search, I had a computer look for solutions with $p,q\le 40$ for $\frac{p}{q}\in\mathbb{Q}$. And there were plenty of non-trivial solutions.
$$\frac{3}{10} + \frac{6}{5} + \frac{3}{1} = \frac{10}{3} + \frac{5}{6} + \frac{1}{3} = \frac{9}{2}$$ $$\frac{10}{3} + \frac{9}{5} + \frac{2}{9} = \frac{3}{10} + \frac{5}{9} + \frac{9}{2} = \frac{241}{45}$$ $$\frac{10}{7} + \frac{4}{7} + \frac{5}{4} = \frac{7}{10} + \frac{7}{4} + \frac{4}{5} = \frac{13}{4}$$ $$\frac{11}{6} + \frac{11}{5} + \frac{3}{10} = \frac{6}{11} + \frac{5}{11} + \frac{10}{3} = \frac{13}{3}$$ $$\frac{4}{15} + \frac{10}{3} + \frac{5}{4} = \frac{15}{4} + \frac{3}{10} + \frac{4}{5} = \frac{97}{20}$$ $$\frac{4}{15} + \frac{12}{5} + \frac{2}{1} = \frac{15}{4} + \frac{5}{12} + \frac{1}{2} = \frac{14}{3}$$ $$\frac{8}{15} + \frac{8}{5} + \frac{6}{5} = \frac{15}{8} + \frac{5}{8} + \frac{5}{6} = \frac{10}{3}$$ $$\frac{3}{20} + \frac{15}{4} + \frac{10}{3} = \frac{20}{3} + \frac{4}{15} + \frac{3}{10} = \frac{217}{30}$$ $$\frac{20}{9} + \frac{5}{12} + \frac{10}{9} = \frac{9}{20} + \frac{12}{5} + \frac{9}{10} = \frac{15}{4}$$ $$\frac{21}{4} + \frac{1}{12} + \frac{7}{1} = \frac{4}{21} + \frac{12}{1} + \frac{1}{7} = \frac{37}{3}$$ $$\frac{26}{9} + \frac{13}{2} + \frac{1}{9} = \frac{9}{26} + \frac{2}{13} + \frac{9}{1} = \frac{19}{2}$$ $$\frac{28}{3} + \frac{2}{21} + \frac{7}{4} = \frac{3}{28} + \frac{21}{2} + \frac{4}{7} = \frac{313}{28}$$ $$\frac{30}{17} + \frac{21}{17} + \frac{10}{21} = \frac{17}{30} + \frac{17}{21} + \frac{21}{10} = \frac{73}{21}$$ $$\frac{7}{33} + \frac{11}{6} + \frac{7}{2} = \frac{33}{7} + \frac{6}{11} + \frac{2}{7} = \frac{61}{11}$$ $$\frac{35}{2} + \frac{20}{7} + \frac{1}{20} = \frac{2}{35} + \frac{7}{20} + \frac{20}{1} = \frac{2857}{140}$$ $$\frac{4}{35} + \frac{28}{5} + \frac{7}{2} = \frac{35}{4} + \frac{5}{28} + \frac{2}{7} = \frac{129}{14}$$ $$\frac{9}{35} + \frac{18}{5} + \frac{7}{6} = \frac{35}{9} + \frac{5}{18} + \frac{6}{7} = \frac{211}{42}$$ $$\frac{36}{7} + \frac{3}{28} + \frac{9}{2} = \frac{7}{36} + \frac{28}{3} + \frac{2}{9} = \frac{39}{4}$$ $$\frac{5}{38} + \frac{20}{3} + \frac{19}{12} = \frac{38}{5} + \frac{3}{20} + \frac{12}{19} = \frac{637}{76}$$ $$\frac{10}{39} + \frac{15}{4} + \frac{13}{12} = \frac{39}{10} + \frac{4}{15} + \frac{12}{13} = \frac{397}{78}$$ $$\frac{39}{10} + \frac{26}{15} + \frac{1}{5} = \frac{10}{39} + \frac{15}{26} + \frac{5}{1} = \frac{35}{6}$$ $$\frac{39}{28} + \frac{3}{28} + \frac{26}{3} = \frac{28}{39} + \frac{28}{3} + \frac{3}{26} = \frac{61}{6}$$ $$\frac{3}{40} + \frac{15}{8} + \frac{12}{1} = \frac{40}{3} + \frac{8}{15} + \frac{1}{12} = \frac{279}{20}$$ $$\frac{11}{40} + \frac{11}{4} + \frac{8}{5} = \frac{40}{11} + \frac{4}{11} + \frac{5}{8} = \frac{37}{8}$$
My attempts:
In addition to the strategies listed in the previous version of this question I found another weird transformation. If we multiply both sides of the original equation through by say $a$, we get $$a^2+a(b+c)=1+a\Big(\frac{1}{b}+\frac{1}{c}\Big)$$ which we can rearrange to $$a^2+av-1=0\quad\text{with}\quad v=b+c-\frac{1}{b}-\frac{1}{c}$$ This quadratic in $a$ turns out to be equivalent to the pythagorean triple $$2^2+v^2=(2a+v)^2$$ And since pythagorean triples are categorized over the rationals $\mathbb{Q}$, I tried matching up the parametrization of pythagorean triples $p^2(2mn, m^2-n^2, m^2+n^2)$ with $(2,v,2a+v)$. But I somehow ended up with the result that $a$ has to be either a perfect square or double a perfect square which is obviously wrong given the solutions above.
My only other lead at the moment is the values of the sums in the solution list. The numerators have some charismatic primes. Also, on a note of interest for the original problem, there don't seem to be any perfect squares in these first few solutions.
And I've had no luck yet finding papers/oeis entries/etc. about this problem so any suggestions are appreciated.