Roots of the equation
$$
x^2+2(a+b+c) x+3 \lambda(a b+b c+c a)=0
$$
are real, so
$$
\begin{aligned}
& {[2(a+b+c)]^2-4(1)[3 \lambda(a b+b c+c a)] \geq 0} \\
& \Rightarrow \quad \lambda \leq \frac{a^2+b^2+c^2+2(a b+b c+c a)}{3(a b+b c+c a)}
\end{aligned}
$$


Now, sum of two roots is greater than thirds
So,
$$
\begin{aligned}
& a < b+c \Rightarrow(a-c) < b \\
& \Rightarrow \quad(a-c)^2 < b^2 \\
&
\end{aligned}
$$
$\Rightarrow \quad a^2+c^2-2 a c < b^2$

Adding Eqs. (ii), (iii) and (iv), we get
$$
\begin{aligned}
& a^2+b^2+c^2 < 2(a b+b c+c a) \\
\Rightarrow \quad & \frac{a^2+b^2+c^2}{a b+b c+c a} < 2
\end{aligned}
$$

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