Let a , b and c be three positive real numbers such that the sum of any two of them is greater than the…

Given,

$x^2+2(a+b+c)x+3\lambda (ab+bc+ca)=0$

For the roots to be real, $D\ge 0$

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

$\Rightarrow (a+b+c{)}^2-3\lambda \left(ab+bc+ca\right)\ge 0$

$\Rightarrow a^2+b^2+c^2+2\sum _{}ab\ge 3\lambda \sum _{}ab$

$\Rightarrow \frac{1}{3}\left\{\frac{{\displaystyle \sum _{}}{a}^2}{{\displaystyle \sum _{}}ab}\right\}+\frac{2}{3}\ge \lambda ...\left(1\right)$

Now, it is given that

$a+b>c ...\left(i\right)$

$b+c>a ...\left(ii\right)$

$c+a>b ...\left(iii\right)$

Multiplying $\left(i\right), \left(ii\right) \& \left(iii\right)$ by $c,a \& b$ respectively and adding we get

$2\left(ab+bc+ca\right)>a^2+b^2+c^2$

$\Rightarrow 2\sum _{}ab>\sum _{}{a}^2$

$\Rightarrow \frac{{\displaystyle \sum _{}}{a}^2}{{\displaystyle \sum _{}}ab}<2 ...\left(2\right)$

From $\left(i\right) \& \left(ii\right)$, we have

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