Question

Let $a,b,c$ be the sides of a triangle where $a\ne b\ne c$ and $\lambda \in R$. If the roots of the equation $x^2+2(a+b+c)x+3\lambda (ab+bc+ca)=0$ are real, then

• A. $\lambda <\frac{4}{3}$
• B. $\lambda <\frac{5}{3}$
• C. $\lambda \in (\frac{4}{3},\frac{5}{3})$
• D. $\lambda \in (\frac{1}{3},\frac{5}{3})$

Answer 1

The correct option is A $\lambda <\frac{4}{3}$

$\therefore a,b,c$ are sides of a triangle and $a\ne b\ne c\therefore |a-b|<|c|\Rightarrow a^2+b^2-2ab<c^2$
Similarly, we have
$a^2+c^2-2bc<a^2;c^2+a^2-2ca<b^2$
On adding, we get
$a^2+b^2+c^2<2(ab+bc+ca)\Rightarrow \frac{a^2+b^2+c^2}{ab+bc+ca}<2...\left(1\right)\therefore$ Roots of the given equation are real
$\therefore (a+b+c{)}^2-3\lambda (ab+bc+ca)\ge 0\Rightarrow \frac{a^2+b^2+c^2}{ab+bc+ca}\ge 3\lambda -2...(2)$
From equations $\left(1\right)$ and $\left(2\right)$, we get
$3\lambda -2<2$
$\Rightarrow \lambda <\frac{4}{3}$