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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!

random guy 2000
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1 Answers1

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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.

Amit Rajaraman
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