<s>[INST] Polynomial Inequality Holds for All Real Numbers x Problem: The quadratic polynomial $P(x),$ with real coefficients, satisfies
$$P(x^3 + x) \ge P(x^2 + 1)$$ for all real numbers $x.$ Find the sum of the roots of $P(x).$
Work: Since $P(x)$ is a quadratic polynomial, let $P(x)=ax^2+bx+c$. Then, we have $$a(x^3+x)^2+b(x^3+x)+c \ge a(x^2+1)^2+b(x^2+1)+c \implies a(x^3+x)^2+b(x^3+x) \ge a(x^2+1)^2+b(x^2+1).$$ Since we need to find the sum of the roots of $P(x)$, we need the value $-\frac{b}{a}$ by Vieta's Formulas. I then tried to manipulate using that form, but got nothing. Can somebody finish off this solution, or take another path of their own to solve this problem? Thanks!
 [/INST] You have
$$(x^3 - x^2 + x - 1) (a(x^3 + x^2 + x + 1) +b) \ge 0.$$
So, for any $x > 1$, $a(x^3+x^2+x+1)+b \ge 0$ and for any $x < 1$, $a(x^3+x^2+x+1)+b \le 0$.
It follows from continuity that $a(1^3+1^2+1+1) + b = 0$, that is, $-b/a = 4$.
You can further check that $x^2-4x$ satisfies the inequality.
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