Find all the integers which are of form $\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}, a,b,c\in \mathbb{N}$, any two of $a,b,c$ are relatively prime. 
I have a question which askes to find all the integers which can be
expressed as
$\displaystyle \tag*{} \dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}$
where $a,b,c\in \mathbb{N} $  and any two of $a,b,c$ are relatively
prime.

My approach:
Since they told any two of $a,b,c$ are relatively prime, so:
$\displaystyle \tag*{} \begin{align} \text{gcd}(a,b) &= 1 \\ \text{gcd}(b,c) &= 1 \\ \text{gcd}(c,a) &= 1\end{align}$
We have:
$\displaystyle \tag*{} \dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c} = \dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{c}$
And we have to prove:
$\displaystyle \tag*{} \frac{a}{b} + \frac{b}{a}+ \frac{c}{a} + \frac{a}{c}+ \frac{b}{c} + \frac{c}{b} \in \mathbb{N} $
We have the following that if $\text{gcd}(m,n)=1$ and $m,n,k \in \mathbb{Z}$ then
$\displaystyle \tag*{} \frac{m}{n} + \frac{n}{m} = k$
has only $1$ positive solution, which is $m=n=1$
So, the only solution I found is $a=b=c=1$.
Is there anything I am missing? Any help would be appreciated, thank you. :)

NB: While I was typing up my proof, @user provided a similar proof, leaving mine essentially a duplicate. However, I'll leave it since I may have written it out in slightly greater detail.
First, we may note that $c$ must divide $a+b$, etc. To see this, write the sum on common denominator
$$
\frac{b+c}a+\frac{a+c}b+\frac{a+b}c
=\frac{bc(b+c)+ac(a+c)+ab(a+b)}{abc}
$$
and note that, for $c$ to divide the numerator, after removing terms that are multiples of $c$, we are left with $c|ab(a+b)$, which forces $c|a+b$. Similarly, $a|b+c$ and $b|a+c$.
We may assume that $a\le b\le c$, where equality is only possible for the number 1. So let's first deal with the cases where equalities are an issue.
First, $a=b=c=1$ is a solution. Ie, $(a,b,c)=(1,1,1)$.
Next, if $a=b=1<c$, we need $c|a+b$, for which the only option is $c=2$. This is also a solution. Ie $(a,b,c)=(1,1,2)$.
So now we are left with the case $a<b<c$. Since $c|a+b$, but $a+b<2c$, we must have $c=a+b$. Next $b|a+c=2a+b$, so $b|2a$; however, $a<b$, so $2a=b$ since $2a$ is a multiple of $b$ but less than $2b$. This also forces $a=1$ since $a$ and $b=2a$ should be relatively prime. This makes $(a,b,c)=(1,2,3)$.
