Question

Let $a,b,c,$ be the sides of a triangle. No two of them are equal 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 and distinct, then

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

Answer 1

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

Given roots of equation

$x^2+2(a+b+c)x+3\lambda (ab+bc+ca)=0$ are real

So, $D\ge 0$

$\Rightarrow B^2-4AC\ge 0$

$\Rightarrow \left[2\right(a+b+c){]}^2-4\times 1\times 3\lambda (ab+bc+ca)\ge 0$

$\Rightarrow 4(a+b+c{)}^2-12\lambda (ab+bc+ca)\ge 0$

$\Rightarrow 4(a+b+c{)}^2\ge 12\lambda (ab+bc+ca)$

$\Rightarrow \lambda \le \frac{4(a+b+c{)}^2}{12(ab+bc+ca)}$

$\Rightarrow \lambda \le \frac{4(a^2+b^2+c^2)}{12(ab+bc+ca)}+\frac{8(ab+bc+ca)}{12(ab+bc+ca)}$

$\Rightarrow \lambda \le \frac{(a^2+b^2+c^2)}{3(ab+bc+ca)}+\frac{2}{3}.....\left(1\right)$

Now, we know that

$|a-b|<c\Rightarrow a^2+b^2-2ab<c^2$

$|b-c|<a\Rightarrow b^2+c^2-2bc<a^2$

$|c-a|<b\Rightarrow c^2+a^2-2ca<b^2$

On adding above, we get

$a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca<a^2+b^2+c^2$

$\Rightarrow a^2+b^2+c^2<2ab+2bc+2ca$

$\Rightarrow \frac{(a^2+b^2+c^2)}{(ab+bc+ca)}<2$

So, equation (1) becomes,

$\Rightarrow \lambda <\frac{2}{3}+\frac{2}{3}$

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