Quadratic expression problem involving sides of a triangle

Question

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 $\lambda$?

Answer 1

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 $(-\infty, \frac43)$.

Answer 2

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 $$R(\lambda)=(- \infty, \frac {(a+b+c)^2}{3(ab+bc+ca)}]$$