Given ax² + bx + c $\ge$ 0; bx² + cx + a > 0 and cx² + ax + b $\ge$ 0 where a $\ne$ b $\ne$ c and a, b, c $\in$R Now $\frac{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}{\mathrm{ab}+\mathrm{bc}+\mathrm{ca}}$ can not take values

Given that a $\ne$ b $\ne$ c and a, b, c $\in$R
Now ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}\ge 0\Rightarrow {\mathrm{b}}^2-4\mathrm{ac}\le 0\text{ and }\mathrm{a}>0$
$$\begin{gathered} {\mathrm{bx}}^2+\mathrm{cx}+\mathrm{a}\ge 0\Rightarrow {\mathrm{c}}^2-4\mathrm{ab}\le 0\text{ and }\mathrm{b}>0 \\ \Rightarrow {\mathrm{cx}}^2+\mathrm{ax}+\mathrm{b}\ge 0\Rightarrow {\mathrm{a}}^2-4\mathrm{bc}\le 0\text{ and }\mathrm{c}>0 \end{gathered}$$
equality can not hold simultaneously $\because \mathrm{a}\ne \mathrm{b}\ne \mathrm{c}$
$$\begin{gathered} \therefore \frac{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}{\mathrm{ab}+\mathrm{bc}+\mathrm{ca}}=4\Rightarrow \because \mathrm{a}\ne \mathrm{b}\ne \mathrm{c} \\ \Rightarrow (\mathrm{a}-\mathrm{b}{)}^2+(\mathrm{b}-\mathrm{c}{)}^2+(\mathrm{c}-\mathrm{a}{)}^2>0 \\ \Rightarrow \frac{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}{\mathrm{ab}+\mathrm{bc}+\mathrm{ca}}>1 \therefore \frac{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}{\mathrm{ab}+\mathrm{bc}+\mathrm{ca}}=\in (1,4) \end{gathered}$$