Calculation of $\lambda$, If $x^2+2(a+b+c)\cdot x+3\lambda \cdot (ab+bc+ca) = 0$ has real roots

Question

Let $a,b,c$ be the sides of a $\triangle$ where $a\neq b\neq c$ and $\lambda\in \mathbb{R}$. If the roots of the equation

$$x^2+2(a+b+c)\cdot x+3\lambda \cdot (ab+bc+ca) = 0$$ are real , Then which one is Right.

$\bf{Options}::$ $\displaystyle (a)\;\; \lambda <\frac{4}{3}\;\;\;\;\;\; (b)\; \lambda >\frac{4}{3}\;\;\;\;\;\; (c)\; \lambda \in \left(\frac{1}{3},\frac{5}{3}\right)\;\;\;\;\;\; (d)\;\; \lambda \in \left(\frac{4}{3},\frac{5}{3}\right)$

$\bf{My\; Try}::$ If given equation has real roots , Then its $\bf{Discriminant}\geq 0$

So $$\displaystyle 4(a+b+c)^2-12\lambda\cdot (ab+bc+ca)\geq 0$$

$$\displaystyle (a+b+c)^2-3\lambda\cdot (ab+bc+ca)\geq 0$$

$$\displaystyle 3\lambda\leq \frac{(a+b+c)^2}{(ab+bc+ca)} = \frac{a^2+b^2+c^2}{(ab+bc+ca)}+2$$

Now I did not understand how can i solve after that

Help Required

Thanks

Answer 1

Use that $(a,b,c)$ forms a triangle, hence $$a \leq b+c \\ b \leq c+a \\ c\leq a+b$$ showing $$2(ab+bc+ca) = a(b+c)+b(c+a)+c(a+b) \geq a^2+b^2+c^2.$$

Thus $\lambda \leq \frac{4}{3}$.