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

Related : http://math.stackexchange.com/questions/280554/let-a-b-and-c-be-the-lengths-of-the-sides-of-an-arbitrary-triangle-pick-ou — lab bhattacharjee, Dec 25 '13 at 14:47

score 3 · Accepted Answer · answered Dec 25 '13 at 14:45

3

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

answered Dec 25 '13 at 14:45

benh

6,605

It seems that you wrote the inequalities backwards... The equilateral triangle doesnt verify that. – Beni Bogosel Dec 25 '13 at 14:47
absolutely right, thanks! – benh Dec 25 '13 at 14:50
Nice solution benh.... – juantheron Dec 25 '13 at 17:05

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

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