Quadratic expression problem involving sides of a triangle If $a,b,c$ be the sides of a triangle where $a\neq b\neq c$ and $\lambda \in R$, if roots of the equation $x^{2} + 2\lgroup a+b+c\rgroup x + 3\lambda\lgroup ab+bc+ca\rgroup =0$ are real, then what is the range of $\lambda$?

First, we incorporate the triangle restriction using the substitution $a = u+v, b = v+w, c = w+u$, where $u, v, w$ are now distinct positive reals.  
So the quadratic we want to have real roots is now:
$$x^2+4(u+v+w)x+3\lambda(u^2+v^2+w^2+3uv+3vw+3wu)$$
The condition for real roots is that the discriminant is non-negative, so we need to have for any allowable choice of $u, v, w$:
$$4(u+v+w)^2 \ge 3\lambda(u^2+v^2+w^2+3uv+3vw+3wu)$$
$$\implies 4(u+v+w)^2 \ge 3\lambda\left((u+v+w)^2+(uv+vw+wu)\right)$$
$$\implies \frac43\frac{(u+v+w)^2}{(u+v+w)^2+(uv+vw+wu)} \ge \lambda$$
So we need to find the maximum possible value for the LHS to get the full range of $\lambda$.
$$\frac43\frac{(u+v+w)^2}{(u+v+w)^2+(uv+vw+wu)}=\frac43\frac1{1+\frac{uv+vw+wu}{(u+v+w)^2}} < \frac43$$
As if $(u, v) \to (0, 1)$, then $uv+vw+wu \to w$ and $(u+v+w) \to (1+w)^2$ so the ratio can be made as small as desired by taking sufficiently large $w$.
Hence the possible range of $\lambda$ is $(-\infty, \frac43)$.
