Jtdylktuy

Let $a$, $b$ and $c$ be $3$ integers with no common factors.

I conjecture that at least one of the three fractions:

$$frac{a}{a+b+c},quadfrac{b}{a+b+c},quadfrac{c}{a+b+c}$$

is irreducible.

I know that they are not necessarily all irreducible.

For instance, taking $(a,b,c)=(2,7,15)$, it is verified that $2$, $7$ and $15$ have no common factors.

We have $a+b+c=24$ so
$$dfrac{2}{2+7+15}=dfrac{2}{24}quad(text{reducible)}$$
$$dfrac{7}{2+7+15}=dfrac{7}{24}quad(text{irreducible)}$$
$$dfrac{15}{2+7+15}=dfrac{15}{24}quad(text{reducible)}$$

I have yet to find any counter-example to the initial statement so I'm pretty certain it must be true. I believe I managed to prove the analogous statement for $2$ integers but I would like to generalize it to $3$, and eventually $n$, integers.

Proof for two integers $a$ and $b$ with no common factors

Suppose $k$ is a factor of $a$ and that $a'$ is an integer such that $a=ka'$. Then:

$$frac{a}{a+b}=frac{ka'}{ka'+b}=frac{ka'}{kleft(a'+dfrac{b}{k}right)}=dfrac{a'}{a'+dfrac{b}{k}}$$

But $k$ is a factor of $a$ and $a$ and $b$ have no common factors by hypothesis, so $k$ does not divide $b$. So $dfrac{b}{k}$ is irreducible. Since $a'$ is an integer and $dfrac{b}{k}$ isn't (assuming $kneq 1$), then $dfrac{a'}{a'+dfrac{b}{k}}$ is not a 'fraction' in the sense that it is not a ratio of two integers, making the original fraction $dfrac{a}{a+b}$ irreducible.

This proves (correct me if I'm wrong) that at least one of the the fractions $dfrac{a}{a+b}$ or $dfrac{b}{a+b}$ is irreducible.

For the case with $3$ integers, a similar argument doesn't seem to work. Taking $(4,5,11)$ as an example:

$$frac{4}{4+5+11}=frac{4}{4left(1+dfrac{5}{4}+dfrac{11}{4}right)}=frac{1}{1+dfrac{16}{4}}=frac{1}{5}$$

the fact that there are now $3$ numbers allows for some fractions to combine into an integer. Because of this, I'm having difficulty proving the statement further.

I'm not particularly good with number theory proofs and I apologize if this is not as elegant as it should be (or worse, if there are mistakes). I'm welcoming any tips and possible improvements. Thanks in advance!

Counterexample: $5,7,58$.
begin{align*}frac{5}{5+7+58} &= frac{5}{70} = frac1{14}\
frac{7}{5+7+58} &= frac{7}{70} = frac1{10}\
frac{58}{5+7+58} &= frac{58}{70} = frac{29}{35}end{align*}

How did I find it? Choose $a$ and $b$ odd and relatively prime, then let $c=2ab-a-b$. This $c$ must be even, which makes $frac{c}{a+b+c}$ reducible.

The smallest counterexample is $(2,3,25)$. An infinite family of counterexamples is $(2,3,25+30n)$:
$$
begin{align*}
frac{a}{a+b+c} &= frac{2}{30+30n} = frac{1}{15+15n}
\ frac{b}{a+b+c} &= frac{3}{30+30n} = frac{1}{10+10n}
\ frac{c}{a+b+c} &= frac{25+30n}{30+30n} = frac{5+6n}{6+6n}
end{align*}
$$

There are several other infinite family of counterexamples.