Let $\Phi$ be a root system of a finite-dimensional semisimple complex Lie
algebra.
Let $a,b,c\in\Phi$ st. $a+b+c=0$. I want to show that $\frac{N_{a,b}}{\langle c,c\rangle}=\frac{N_{b,c}}{\langle a,a\rangle}=\frac{N_{c,a}}{\langle b,b\rangle}$.
Here we consider the inner product defined on the dual space ($\mathfrak{h}^*$) of a fixed Cartan subalgebra of our Lie Algebra. $N_{i,j}$ is the structure constants, ie. the coefficient of a given Lie bracket relation for Cartan Weyl basis elements, specifically $[e_i,e_j]=N_{i,j}e_{i+j}$ if $i+j\in\Phi$.
My thought is to use the Lie bracket relations of the Cartan Weyl basis, together with the jacobi identity on $e_a,e_b,e_c$. This gives
$$N_{a,b}h_c+N_{b,c}h_a+N_{c,a}h_b=0$$
or equivalently
$$\frac{N_{a,b}}{\langle c,c\rangle}t_c+\frac{N_{b,c}}{\langle a,a\rangle}t_a+\frac{N_{c,a}}{\langle b,b\rangle}t_b=0$$
Since $[e_i,e_{-i}]=h_i=\frac{2}{\langle i,i\rangle}t_i$ where $t_i$ corresponds to the image of $i$ under the bijection $\psi:\mathfrak{h}^*\rightarrow \mathfrak{h}, i\mapsto t_i$.
Here the $h_i$'s are elements of the Cartan Weyl basis.
From $N_{a,b}h_c+N_{b,c}h_a+N_{c,a}h_b=0$, I could use that $h_a+h_b+h_c=0$ (as $a+b+c=0$) and the linearly independence of the $h_i$'s to conclude $N_{a,b}=N_{b,c}=N_{c,a}$. So this is pretty close to what I want.
Now I haven't managed to get any further. I know that $t_a, t_b, t_c$ spans $\mathfrak{h}$ and not much more. I was thinking, that since $\psi$ is a bijection and $a+b+c=0$ then $t_a+t_b+t_c=0$ (however I'm not sure if this is actually true?), and one could maybe use this to conclude something.
Best Answer
It is not in general true that $a+b+c=0$ implies $h_a + h_b + h_c=0$ -- in fact, the map $r \mapsto h_r$ can fail to be a morphism of root systems, in particular, to be additive. Namely, the $h_a$ are the coroots, i.e. come from the dual root system, which can differ from the original root system.
However, the map $r \mapsto t_r$ indeed is an isomorphism of root systems, hence $a+b+c=0$ does imply that $t_a+t_b+t_c = 0$ and then indeed the equation follows as outlined by you. More precise than "linear independence" would be to say though that the space spanned by $a,b,c$ (or $t_a, t_b, t_c$) has (for example) $a,b$ as a basis, which gives identity of the coefficients.
Note further that in the case of simply laced root systems, the above distinction does not matter since all roots have the same length or squared length $\langle a, a \rangle=\langle b, b \rangle=\langle c, c \rangle$ anyway. As soon as they don't, though, indeed by two-dimensionality of the respective space the equation $N_{a,b}=N_{b,c}=N_{c,a}$ would actually contradict the one you're looking for, so it's good it is wrong.