Problem
ALG-B3-M02-P022 A quadratic on a half-line
A quadratic polynomial \(p(x)\) has two distinct real roots and satisfies for all real \(u,v\): \[p(u^2+4v^2)\ge p(4uv).\] Prove that at least one root of \(p\) is negative.
Hint 1. Show that for any \(0\le r\le q\), it is possible to represent \(r=4uv\), \(q=u^2+4v^2\).
Hint 2. Thus \(p\) is nondecreasing on the nonnegative half-line.
Put \(a=u\), \(b=2v\). The condition becomes \(p(a^2+b^2)\ge p(2ab)\). For any \(0\le r\le q\), there exist nonnegative \(a,b\) with \(2ab=r\), \(a^2+b^2=q\): take \(a=\frac{\sqrt{q+r}+\sqrt{q-r}}{2}\), \(b=\frac{\sqrt{q+r}-\sqrt{q-r}}{2}\). Hence \(p(q)\ge p(r)\) whenever \(0\le r\le q\), so \(p\) is nondecreasing on \([0,\infty)\). If both roots were nonnegative, a quadratic with two distinct roots would change sign between and after them, and could not be nondecreasing on all of \([0,\infty)\). Thus at least one root is negative.
A. Source analysis. Main objects: a quadratic polynomial and comparison of values at u^2+4v^2 and 4uv.
B. Insufficient first move. Substituting u=0 or u=v gives only partial information.
C. Hidden observation. The pair (u^2+4v^2, 4uv) encodes any 0<=r<=q.
D. Required move. Replace 2v by a new variable and prove monotonicity on the nonnegative half-line.
E. Number of ideas. Three ideas: variable normalization, covering all pairs r<=q, and a graph contradiction for two nonnegative roots.
F. Difficulty justification. Regional level 7: it requires a hidden parametrization, not a single substitution.
G. Why it is not one-step. It is not one-step: after substitution one must convert the inequality to monotonicity and then to root behavior.