Let $a, b, c$ be the sides of a triangle where $a\ne c$ 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 

Since roots are real, so 

$\begin{array}{l} D>0\\ \Rightarrow 4(a+b+c{)}^2-12\lambda (ab+bc+ca)>0\\ \Rightarrow (a+b+c{)}^2-3\lambda (ab+bc+ca)>0\\ \Rightarrow (a+b+c{)}^2+3\lambda (ab+bc+ca)\\ \Rightarrow \left(a^2+b^2+c^2\right)>(3\lambda -2)(ab+bc+ca)\end{array}$

$\begin{array}{l}\Rightarrow (3\lambda -2)<\frac{\left(a^2+b^2+c^2\right)}{(ab+bc+ca)}\\ \Rightarrow (3\lambda -2)<2\end{array}$

$\Rightarrow \lambda <\frac{4}{3}$  $$\begin{gathered} \left[\text{ since }\cos A=\frac{b^2+c^2-a^2}{2bc}\le 1\right.b^2+c^2-a^2\le 2bc \\ Similarly, c^2+a^2-b^2\le 2ac and a^2+b^2-c^2\le 2ab \\ Thus, a^2+b^2-c^2\le 2 (ab+bc+ca) \\ \left.\Rightarrow \frac{\left(a^2+b^2+c^2\right)}{(ab+bc+ca)}\le 2\right] \end{gathered}$$