A polynomial with all roots real and Vieta

Vieta. $p=-(a+b+c)$ and $q=ab+bc+ca$. Hence
$$p^2-3q=(a+b+c)^2-3(ab+bc+ca)=a^2+b^2+c^2-ab-bc-ca.$$

Sum of squares.
$$a^2+b^2+c^2-ab-bc-ca=\frac12\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\ge0.$$

Equality. It requires $a=b$, $b=c$, $c=a$, i.e. $a=b=c$.

Remark. Geometrically, $p^2\ge3q$ says that the derivative $3x^2+2px+q$ has a real root: this is Rolle's theorem applied between the real roots of the cubic. The condition $p^2-3q\ge0$ is exactly the non-negativity of the discriminant of the derivative.

Lesson. A relation between coefficients of a polynomial with real roots usually comes from a sum of squares of differences of the roots, via Vieta's formulas.