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

Answer 1

The correct option is A

$\lambda <\frac{4}{3}$

Since, roots are real, therefore $D\ge 0$
$\Rightarrow 4(a+b+c{)}^2-12\lambda (ab+bc+ca)\ge 0$
$\Rightarrow (a+b+c{)}^2\ge 3\lambda (ab+bc+ca)$
$\Rightarrow a^2+b^2+c^2\ge (ab+bc+ca)(3\lambda -2)$
$\Rightarrow 3\lambda -2\le \frac{a^2+b^2+c^2}{ab+bc+ca}$
Also, $cosA=\frac{b^2+c^2-a^2}{2bc}<1$
$\Rightarrow b^2+c^2-a^2<2bc$
Similarly, $c^2+a^2-b^2<2ca$
And $a^2+b^2-c^2<2ab$
$\Rightarrow a^2+b^2+c^2<2(ab+bc+ca)$
$\Rightarrow \frac{a^2+b^2+c^2}{ab+bc+ca}<2$
From Eqs. (i) and (ii) we get
$3\lambda -2<2\Rightarrow \lambda <\frac{4}{3}$