Both the roots of the given equation

$(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})+(\mathrm{x}-\mathrm{b})(\mathrm{x}-\mathrm{c})+(\mathrm{x}-\mathrm{c})(\mathrm{x}-\mathrm{a})=0$ are always

Given equation

$(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})+(\mathrm{x}-\mathrm{b})(\mathrm{x}-\mathrm{c})+(\mathrm{x}-\mathrm{c})(\mathrm{x}-\mathrm{a})=0$

can be re-written as

$$\begin{gathered} 3{\mathrm{x}}^2-2(\mathrm{a}+\mathrm{b}+\mathrm{c})\mathrm{x}+(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})=0 \\ ∆=4\{{(\mathrm{a}+\mathrm{b}+\mathrm{c})}^2-3(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\left)\right\} (\because {\mathrm{b}}^2-4\mathrm{ac}=∆) \\ =4({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ac}) \\ =2\{{(\mathrm{a}-\mathrm{b})}^2+{(\mathrm{b}-\mathrm{c})}^2+{(\mathrm{c}-\mathrm{a})}^2\}\ge 0 \end{gathered}$$

Hence both roots are always real.