Conjecture: if $a$, $b$ and $c$ have no common factors, dividing each of them by their sum yields at least one irreducible fraction 
Let $a$, $b$ and $c$ be $3$ integers with no common factors.
I conjecture that at least one of the three fractions:
$$\frac{a}{a+b+c},\quad\frac{b}{a+b+c},\quad\frac{c}{a+b+c}$$
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:
$$\frac{a}{a+b}=\frac{ka'}{ka'+b}=\frac{ka'}{k\left(a'+\dfrac{b}{k}\right)}=\dfrac{a'}{a'+\dfrac{b}{k}}$$
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 $k\neq 1$), 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:
$$\frac{4}{4+5+11}=\frac{4}{4\left(1+\dfrac{5}{4}+\dfrac{11}{4}\right)}=\frac{1}{1+\dfrac{16}{4}}=\frac{1}{5}$$
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!

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.
