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 $\lambda$?

From the quadratic formula $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ from your formula we have $$a=1,b=2(a+b+c),c=3\lambda (ab+bc+ca)$$ Subbing in $$x= \frac{-2(a+b+c) \pm \sqrt{(2(a+b+c))^2-12\lambda (ab+bc+ca)}}{2}$$ The radical is what deterimines if we will have real roots, meaning that we need $$(2(a+b+c))^2-12\lambda (ab+bc+ca) \ge 0\\\lambda \le\frac {(a+b+c)^2}{3(ab+bc+ca)}$$ This will give you a range of $$R(\lambda)=(- \infty, \frac {(a+b+c)^2}{3(ab+bc+ca)}]$$
