Mathematical Reflections 2017, Issue 3 - Problem U412

Problem:
Let $P(x)$ be a monic polynomial with real coefficients, of degree $n$, which has $n$ real roots. Prove that if
$$P(c) \leq \left(\dfrac{b^2}{a}\right)^n,$$
then $P(ax^2+2bx+c)$ has at least one real root.

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
If $P(x)$ is an $n$-th degree monic polynomial with real coefficients and with $n$ real roots $\alpha_1,\ldots,\alpha_n$, then
$$P(x)=(x-\alpha_1)(x-\alpha_2)\cdot \ldots \cdot (x-\alpha_n).$$
Hence, $$P(c)=(c-\alpha_1)(c-\alpha_2)\cdot \ldots \cdot (c-\alpha_n) \leq \left(\dfrac{b^2}{a}\right)^n$$
and $$P(ax^2+2bx+c)=(ax^2+2bx+c-\alpha_1)(ax^2+2bx+c-\alpha_2)\cdot \ldots \cdot (ax^2+2bx+c-\alpha_n).$$
Assume by contradiction that $P(ax^2+2bx+c)$ has no real roots. Then, each factor has negative discriminant, i.e.
$$\begin{array}{lll} b^2-a(c-\alpha_1) & < & 0 \\ b^2-a(c-\alpha_2) & < & 0 \\ \vdots & \vdots & \vdots \\ b^2-a(c-\alpha_n) & < & 0. \end{array} \iff \begin{array}{lll} b^2 & < & a(c-\alpha_1) \\ b^2 & < & a(c-\alpha_2) \\ \vdots & \vdots & \vdots \\ b^2 & < & a(c-\alpha_n). \end{array}$$
Multiplying side by side all these inequalities, we get $(b^2)^n < a^n(c-\alpha_1)(c-\alpha_2)\cdot \ldots \cdot (c-\alpha_n)$, i.e.
$$\left(\dfrac{b^2}{a}\right)^n < (c-\alpha_1)(c-\alpha_2)\cdot \ldots \cdot (c-\alpha_n),$$ contradiction.