If \(a, b\) and \(c\) are the sides of a \(\triangle A B C\) such that \(x^2-2(a+b+c) x+3 \lambda(a b+b c+c a)=0\) has real roots, then

Since, roots are real, \(D \geq 0\)
\[
\begin{array}{ll}
\Rightarrow & 4(a+b+c)^2-12 \lambda(a b+b c+c a) \geq 0 \\
\Rightarrow & (a+b+c)^2 \geq 3 \lambda(a b+b c+c a) \\
\Rightarrow & a^2+b^2+c^2 \geq(a b+b c+c a)(3 \lambda-2) \\
\Rightarrow & 3 \lambda-2 \leq \frac{a^2+b^2+c^2}{a b+b c+c a} \\
\text { where, } a^2+b^2 \geq 2 a b, b^2+c^2 \geq 2 b c \text { and } c^2+a^2 \geq 2 c a \\
\Rightarrow & a^2+b^2+c^2 \geq(a b+b c+c a) \\
\therefore & \frac{a^2+b^2+c^2}{a b+b c+c a} \geq 1
\end{array}
\]
Also, \(\quad \cos A=\frac{b^2+c^2-a^2}{2 b c} < 1\)
\[
\Rightarrow \quad b^2+c^2-a^2 < 2 b c
\]
Similarly, \(c^2+a^2-b^2 < 2 c a\) and \(a^2+b^2-c^2 < 2 a b\)
\[
\Rightarrow a^2+b^2+c^2 < 2(a b+b c+c a) \text { or } \frac{a^2+b^2+c^2}{a b+b c+c a} < 2
\]
From Eqs. (i), (ii) and (iii)
\[
\begin{array}{lrl}
\therefore & 3 \lambda-2 & < 2 \\
\Rightarrow & \lambda < \frac{4}{3}
\end{array}
\]