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

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

Since,                           $|a-b|<c$

$\begin{array}{cc}\Rightarrow & a^2+b^2-2ab<c^2\\ \Rightarrow & |b-c|<a\\ & b^2+c^2-2bc<a^2\\ & \Rightarrow |c-a|<b\\ \Rightarrow & c^2+a^2-2ca<b^2\end{array}$

From Eqs. (i), (ii) and (iii), we get

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

From Eqs. (i) and (iv), we get 

                          $3\lambda -2<2\Rightarrow \lambda <\frac{4}{3}$