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
• 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

The Correct Option is 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, $cosA=\frac{b^2+c^2-a^2}{2bc}<1 \Rightarrow b^2+c^2-a^2<2bc$
Similarly, $\hspace15mm\, \, \, c^2+a^2-b^2<2ca$
and $\hspace25mm\, \, \, a^2+b^2-c^2<2ab$
$\Rightarrow\hspace25mm a^2+b^2+c^2<2(ab+bc+ca)$
$\Rightarrow\hspace25mm\frac{a^2+b^2+c^2}{ab+bc+ca}<2\, \, \, \, \, \, \, \, \, \, \, ...(ii)$
From Eqs. (i) and (ii), we get
$\hspace25mm 3\lambda-2<2 \Rightarrow \lambda