<s>[INST] Find integer in the form: $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$ 
Let $a,b,c \in \mathbb N$ find integer in the form: $$I=\frac{a}{b+c}
 + \frac{b}{c+a} + \frac{c} {a+b}$$

Using Nesbitt's inequality: $I \ge \frac 32$
I am trying to prove $I \le 2$ to implies there $\nexists \ a,b,c$ such that $I\in \mathbb Z$: $$I\le 2 \\ \iff c{a}^{2}+3\,acb+{a}^{3}+{a}^{2}b+a{b}^{2}+{b}^{3}+c{b}^{2}+{c}^{2}b+{c
}^{3}+{c}^{2}a-2\, \left( b+c \right)  \left( c+a \right)  \left( a+b
 \right) \le 0 \\ \iff {a}^{3}+{b}^{3}+{c}^{3}\leq c{a}^{2}+{a}^{2}b+a{b}^{2}+c{b}^{2}+{c}^{2
}a+{c}^{2}b+abc
$$
and stuck.
EDIT: Look like prove $I \le 2$ not a good thinking :P
 [/INST] This not a complete solution, just a 'case' of a problem.
When $\gcd(a+b,a+c,b+c)=1 \implies \text { } a,b,c  \text { are not all odd or not all even}$
Your expression is 
$\dfrac{a(c+a)(a+b)+b(b+c)(a+b)+c(a+c)(b+c)}{(a+b)(b+c)(c+a)}$
Denoting $a+b=k, b+c=l$ and $c+a=m$
$$klm |akm+blk+cml \implies a=p_1l, b=p_2m, c=p_3k$$
$a=p_1l, b=p_2m,$ and $ c=p_3k$ is not possible. So, we have no solution if we consider $\gcd(a+b,a+c,b+c)=1$.
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