Crossing the $$ x $$-axis

Let \(f(x)\), \(g(x)\), and \(h(x)\) denote the polynomials defined by

\(\begin{align*} f(x) & = (x - a)(x - b) - (x - c), \\ g(x) & = (x - b)(x - c) - (x - a), \\ h(x) & = (x - c)(x - a) - (x - b). \end{align*} \\)

Note that

\(\begin{align*} f(x) + g(x) & = (x - b)(2x - a - c) - (2x - a - c) \\ & = (2x - a - c)(x - b - 1). \end{align*} \\)

This implies that \(f(x) + g(x)\) admits a real root. Hence, at least one of \(f(x)\) and \(g(x)\) is non-positive at some real number. If one of them admits a real root, then we are done. Otherwise, at least one of them is negative at some real number. Also note that if \(k\) is a real number satisfying \(k \geq \vert a \vert + \vert b \vert + \vert c \vert + 2\), then

\(\begin{align*} f(k + \vert a \vert + \vert b \vert) & = (k + \vert a \vert + \vert b \vert + a)(k + \vert a \vert + \vert b \vert + b) - (k + \vert a \vert + \vert b \vert + c) \\ & \geq k^2 - k - \vert a \vert - \vert b \vert - \vert c \vert \\ & \geq k - \vert a \vert - \vert b \vert - \vert c \vert \\ & > 0 \end{align*} \\)

holds, and similarly, \[ g(k + \vert a \vert + \vert b \vert) > 0 \] holds too. By the intermediate value theorem, at least one of \(f(x)\) and \(g(x)\) admits a real root. Interchanging the roles of \(f\), \(g\), and \(h\), it follows that at least one of any two of the polynomials \(f\), \(g\), and \(h\) admits a real root. Therefore, at least two of the three equations have real solutions for \(x\).

Consider the three quadratic polynomials defined by

\(\begin{align*} f(x) & = x^2 + (a - b)x + (b - c),\\ g(x) & = x^2 + (b - c)x + (c - a),\\ h(x) & = x^2 + (c - a)x + (a - b). \end{align*} \\)

Note that \[ f(0) + g(0) + h(0) = 0, \] which implies that at least one of \(f(0)\), \(g(0)\), \(h(0)\) is non-positive. Without loss of generality, assume that \(f(0) \leq 0\). If \(f(0)\) is zero, then \(x=0\) is a root of \(f(x)\). If \(f(0) < 0\), then noting that

\(\begin{align*} & f(2 \vert a - b \vert + 2\vert b - c\vert + 1)\\ & = \frac{1}{2} (2 \vert a - b \vert + 2\vert b - c\vert + 1)^2 + (a - b)(2 \vert a - b \vert + 2\vert b - c\vert + 1)\\ & \quad + \frac{1}{2} (2 \vert a - b \vert + 2\vert b - c\vert + 1)^2 + (b - c) \\ & = \frac{1}{2} (2 \vert a - b \vert + 2\vert b - c\vert + 1) \left( 2 \vert a - b \vert + 2\vert b - c\vert + 1 - 2(a - b) \right) \\ & \quad + \frac{1}{2} (2 \vert a - b \vert + 2\vert b - c\vert + 1)^2 + (b - c) \\ & \geq 1 \end{align*} \\)

holds, it follows from the intermediate value theorem that \(f(x)\) has a real root.