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

$D\ge 0$ of Equation : $x^2+2(a+b+c)x+3\lambda (ab+bc+ca)=0$

$\Rightarrow 4(a+b+c{)}^2-12\lambda (ab+bc+ca)\ge 0$ [where $D=b^2-4ac$]

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

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

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

since in a triangle

$|b-a|<c=>a^2+b^2-2ab<c^2$ $\dots \left(1\right)$

$|b-c|<a=>b^2+c^2-2cb<a^2$ $\dots \left(2\right)$

$|c-a|<b=>c^2+a^2-2ca<b^2$ $\dots \left(3\right)$

Add $\left(1\right)+\left(2\right)+\left(3\right)$

we get

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

$(a^2+b^2+c^2)<2(ab+bc+ca)$

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

Put this in $\left(I\right)$ we get

$\frac{2}{3}+\frac{2}{3}>\frac{(a^2+b^2+c^2)}{3(ab+bc+ca)}+\frac{2}{3}\ge \lambda$

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