Ask for the rational roots of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4.$ I consider the problem: ask the rational roots of $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4.$$
I try to use the theory of ellipic curves (If we have two rational roots, we could have the third collinear one) and calculte the genus of $\Sigma_{p}:=\{[a,b,c]: p(a,b,c)=0\}$, where $$p(a,b,c)=a(a+c)(a+b)+b(b+c)(b+a)+c(c+a)(c+b)-4(a+b)(a+c)(b+c).(*)$$
How to get the genus? I know I should use:

The Riemann-Hurwitz theorem: $$2g(\Sigma)-2=B_{p}(f)-2\deg(f),$$
  where $g(\Sigma)$ is the genus of Riemann surface $\Sigma$ and $B_{p}(f)$ branch numbers of $f$ at $p$.

I try to bring $(*)$ into the form $y^2=x(x-1)(x-\lambda), \lambda \ne0,1,$  since I know the genus is one. 
Maybe help?
$p(a,b,c)=a^3+b^3+c^3-3a^2b-3ab^2-3a^2c-3ac^2-3cb^2-3bc^2.$
 A: There is one idea. To search for the solution of the equation.
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=q$$
If we know any solution $(a,b,c)$ of this equation. Then it is possible to find another $(a_2, b_2, c_2)$.  Make such a change.
$$y=(a+b+2c)(q(a+b)-c)-(a+b)^2-(a+c)(b+c)$$
$$z=(a+2b+c)(2b-qa-(q-1)c)+(b+2a+c)(2a-qb-(q-1)c)$$
Then the following solution can be found by the formula.
$$a_2=((5-4q)c-(q-2)(3b+a))y^3+((5-4q)b+4(1-q)c)zy^2+$$
$$+(3c+(q-1)(a-b))yz^2-az^3$$
$$b_2=((5-4q)c-(q-2)(3a+b))y^3+((5-4q)a+4(1-q)c)zy^2+$$
$$+(3c+(q-1)(b-a))yz^2-bz^3$$
$$c_2=2(q-2)cy^3+3(2-q)(a+b)zy^2+$$
$$+((5-4q)(a+b)+2(1-q)c)yz^2+(2c-(q-1)(a+b))z^3$$
A: This is a special case of MO question "Estimating the size of solutions of a diophantine equation". Using the results there, the elliptic curve $\,E: y^2 = x^3 + 109 x^2 + 224 x\,$ corresponds to your $\,\Sigma_p.\,$ The curve has $j$-invariant $1408317602329/2153060$ and $\,\lambda = (845 + 109 \sqrt{65})/1690\,$ gives the Legendre form. The curve $\,E\,$ has a torsion point $\,(56,728)\,$ of order $6$ and a point $\,(-100,260)\,$ of infinite order.
