MOPSS, 29th November 2025

Note that

\(\begin{align*} P(x) & = a_0 x^n + a_1 x^{n - 1} + \dots + a_{n - 1}x + a_n \\ & = a_0 (x^n - x^{n - 1}) + (a_0 + a_1)(x^{n - 1} - x^{n - 2}) \\ & \quad + \dots + (a_0 + a_1 + \dots + a_{n - 1})(x - 1) + (a_0 + a_1 + \dots + a_n) \end{align*} \\)

holds, which implies that

\(\begin{align*} & P(x) - m x^n \\ & = a_0 (x^n - x^{n - 1}) + (a_0 + a_1)(x^{n - 1} - x^{n - 2}) \\ & \quad + \dots + (a_0 + a_1 + \dots + a_{n - 1})(x - 1) + (a_0 + a_1 + \dots + a_n) \\ & \quad - m (x^n - x^{n - 1}) - m (x^{n - 1} - x^{n - 2}) - \dots - m (x - 1) - m \\ & = (a_0 - m)(x^n - x^{n - 1}) + (a_0 + a_1 - m)(x^{n - 1} - x^{n - 2}) \\ & \quad + \dots + (a_0 + a_1 + \dots + a_{n - 1} - m)(x - 1) + (a_0 + a_1 + \dots + a_n - m) . \end{align*} \\)

It follows that for any \(x \geq 1\), \[ P(x) \geq m x^n . \]

We provide an alternative solution using mathematical induction on \(n\).

Note that

\(\begin{align*} & \frac{a_1}{1} + \frac{a_2}{1 + a_1} + \frac{a_3}{1 + a_1 + a_2} + \dots + \frac{a_n}{1 + a_1 + a_2 + \cdots + a_{n-1}} \\ & = \frac{1 + a_1}{1} + \frac{1 + a_1 + a_2}{1 + a_1} + \frac{1 + a_1 + a_2 + a_3}{1 + a_1 + a_2} \\ & \quad + \dots + \frac{1 + a_1 + a_2 + \cdots + a_{n}}{1 + a_1 + a_2 + \cdots + a_{n - 1}} - n \\ & \geq n \sqrt[n]{1 + a_1 + a_2 + \cdots + a_n} - n\\ & = n \sqrt[n]{2^n} - n \\ & = n. \end{align*} \\)

This shows that the minimum possible value of the given expression is at least \(n\). Also note that if \(a_i = 2^{i - 1}\) for all \(1 \leq i \leq n\), then \(a_1 + a_2 + \cdots + a_n = 2^n - 1\) holds, and

\(\begin{align*} & \frac{a_1}{1} + \frac{a_2}{1 + a_1} + \frac{a_3}{1 + a_1 + a_2} + \dots + \frac{a_n}{1 + a_1 + a_2 + \cdots + a_{n-1}} \\ & = \frac{1 + a_1}{1} + \frac{1 + a_1 + a_2}{1 + a_1} + \frac{1 + a_1 + a_2 + a_3}{1 + a_1 + a_2} \\ & \quad + \dots + \frac{1 + a_1 + a_2 + \cdots + a_{n}}{1 + a_1 + a_2 + \cdots + a_{n - 1}} - n \\ & = 2n - n \\ & = n. \end{align*} \\)

This proves that the minimum possible value of the given expression is equal to \(n\).

Note that

\(\begin{align*} (a + b)^3 & = a(a^2 + 3ab + 3b^2) + b^3 \\ & = a(a^2 + 3ab + 3b^2 - 1) + a + b^3 \end{align*} \\)

holds. Since \(a + b^3\) is divisible by \(a^2 + 3ab + 3b^2 - 1\), it follows that \((a + b)^3\) is also divisible by \(a^2 + 3ab + 3b^2 - 1\). Assume for the sake of contradiction that \(a^2 + 3ab + 3b^2 - 1\) is not divisible by the cube of an integer greater than \(1\). It follows that \(a^2 + 3ab + 3b^2 - 1\) divides \((a + b)^2\). This implies that \[ a^2 + 2ab + b^2 \geq a^2 + 3ab + 3b^2 - 1, \] which simplifies to \[ ab + 2b^2 \leq 1, \] which is impossible since \(a\) and \(b\) are positive integers. This proves that \(a^2 + 3ab + 3b^2 - 1\) is divisible by the cube of an integer greater than \(1\).

Assume that there exist nonzero reals \(a, b, c\) satisfying the given condition. Let \(n > 3\) be a positive integer, and let \(P_{n}(x) = x^{n} + \dots + a x^{2} + b x + c\) be a polynomial with exactly \(n\) integral roots, which we denote by \(r_{1}, r_{2}, \dots, r_{n}\), counted with multiplicities. Since \(c\) is nonzero, none of the roots \(r_{1}, r_{2}, \dots, r_{n}\) is equal to zero. It follows that \[ \frac{1}{r_1}, \frac{1}{r_2}, \dots, \frac{1}{r_n} \] are the roots of the polynomial \[ Q_n(x) = x^n P\left( \frac{1}{x} \right) = c x^n + b x^{n-1} + a x^{n-2} + \dots + x + 1. \] By Vieta’s formulas, we obtain \[ \frac{1}{r_1} + \dots + \frac{1}{r_n} = - \frac{b}{c} , \quad \sum_{1 \leq i < j \leq n} \frac{1}{r_i r_j} = \frac{a}{c} . \] This implies that \[ \left( \frac{1}{r_1} + \dots + \frac{1}{r_n} \right)^2 - 2 \sum_{1 \leq i < j \leq n} \frac{1}{r_i r_j} = \frac{b^2 - 2 a c}{c^2} , \] and hence \[ \frac{1}{r_1^2} + \dots + \frac{1}{r_n^2} = \frac{b^2 - 2 a c}{c^2} . \] Also note that \[ r_1 r_2 \dots r_n = (-1)^{n} c . \] It follows that \[ b^2 - 2ac \geq \frac{r_1^2 r_2^2 \dots r_n^2}{r_1^2} + \dots + \frac{r_1^2 r_2^2 \dots r_n^2}{r_n^2} \geq n. \] This shows that \(n \leq b^2 - 2ac\) for any integer \(n > 3\), which is a contradiction. Therefore, there do not exist nonzero reals \(a, b, c\) satisfying the given condition.

If one of the polynomials \(f(x)\), \(g(x)\) has no real roots, then the conclusion follows immediately. Thus, we may assume that both \(f(x)\) and \(g(x)\) have real roots. Denote the roots of \(f(x)\) by \(\alpha_1, \alpha_2\), and the roots of \(g(x)\) by \(\beta_1, \beta_2\). We need to show that none of \(\alpha_1, \alpha_2\) lies in \(f(\mathbb{R})\), or none of \(\beta_1, \beta_2\) lies in \(g(\mathbb{R})\).

We claim that if \(\alpha_2 \leq \beta_2\), then none of \(\alpha_1, \alpha_2\) lies in \(f(\mathbb{R})\). To prove the Claim, let us assume that \(\alpha_1 \leq \beta_2\). If some of \(\alpha_1, \alpha_2\) lies in \(f(\mathbb{R})\), then noting that \(f\) is a monic quadratic polynomial, it follows from the intermediate value theorem that \(\beta_2\) lies in \(f(\mathbb{R})\), which shows that the equation \(g(f(x)) = 0\) has a real root, a contradiction. This proves the claim.

If \(\alpha_2 \leq \beta_2\), then by the Claim, the equation \(f(f(x)) = 0\) has no real roots. Similarly, if \(\beta_2 \leq \alpha_2\), then the equation \(g(g(x)) = 0\) has no real roots. This completes the proof.

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.