Question

If $a, b, c$ are the sides of a triangle $ABC$ such that $x^2 2(a+b+c)x+3\lambda(ab+bc+ca)=0$ has real roots, then

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

Answer 1

Answer: (A)

Since, roots are real, therefore $D\ge0$
$\Rightarrow 4(a+b+c)^2-12\lambda(ab+bc+ca)\ge0$
$\Rightarrow (a+b+c)^2\ge3\lambda(ab+bc+ca)$
$\Rightarrow (a^2+b^2+c^2)\ge(ab+bc+ca)(3\lambda-2)$
$\Rightarrow 3\lambda-2 \le \frac{a^2+b^2+c^2}{ab+bc+ca} ...(i)$
Also, $\cos A=\frac{b^2+c^2-a^2}{2bc} < 1 \Rightarrow b^2+c^2-a^2 < 2bc$
Similarly, $c^2+a^2-b^2 < 2ca$
and $ a^2+b^2-c^2 < 2ab$
$\Rightarrow a^2+b^2+c^2 < 2(ab+bc+ca)$
$\Rightarrow \frac{a^2+b^2+c^2}{ab+bc+ca} < 2 ...(ii)$
From Eqs. (i) and (ii), we get
$ 3\lambda-2 < 2 \Rightarrow \lambda < \frac{4}{3}$