Quadratic polynomials

Without loss of generality, let us consider the case that the equations \[ (x - a)(x - b) = x - c, (x - b)(x - c) = x - a \] do not any real root. It follows that their discriminants are negative, that is, \[ (a + b + 1)^2 - 4 (ab + c), (b + c + 1)^2 - 4(bc + a) \] are negative. Note that

\(\begin{align*} (a + b + 1)^2 - 4(ab + c) & = (a - b)^2 + 1 + 2(a + b - 2c)\\ & = (b - a + 1)^2 + 4(a - c),\\ (b + c + 1)^2 - 4(bc + a) & = (b - c)^2 + 1 + 2(b + c - 2a)\\ & = (b - c + 1)^2 + 4(c - a). \end{align*} \\)

Since \(a\) and \(c\) are distinct real numbers, one of \(a - c\) and \(c - a\) is positive. Thus, the discriminant of one of the equations \[ (x - a)(x - b) = x - c, (x - b)(x - c) = x - a \] is non-negative, which is a contradiction. Therefore, at least two of the three equations have real solutions.

Note that the sum of the discriminants of the quadratic polynomials \[ x^2 + (a - b)x + (b - c), x^2 + (b - c)x + (c - a), x^2 + (c - a)x + (a - b) \] are equal to

\(\begin{align*} & (a - b)^2 - 4(b - c) + (b - c)^2 - 4(c - a) + (c - a)^2 - 4(a - b) \\ & = (a - b)^2 + (b - c)^2 + (c - a)^2 , \end{align*} \\)

which is non-negative, and hence, at least one of the discriminants is non-negative. This implies that at least one of the three equations has a real root.