A quadratic polynomial P has the property that $P(x^3+x)\ge P(x^2+1)$ for all real numbers x. Find the sum of the roots of P.

$\text{Let}P\left(x\right)=a{x}^2+bx+c\text{. Then}$
$P(x^3+x)-P(x^2+1)=a{x}^2{(x^2+1)}^2+bx(x^2+1)-a{(x^2+1)}^2-b(x^2+1)=$

$a{(x^2+1)}^2(x^2-1)+b(x-1)(x^2+1)=(x-1)(x^2+1)\left(a\right(x+1)(x^2+1)+b).$

Since this is always nonnegative, we have

$(x-1)\left(a\right(x+1)(x^2+1)+b)\ge 0$
for all x. Taking x=1+u, we deduce that $a(u+2)(1+(1+u{)}^2)+b$ has the same sign as u for all  $u\ne 0$ When u is negative and very close to 0 , the number $a(u+2)(1+(1+u{)}^2)+b$ is very close to $4a+b, so 4a+b$ must be nonpositive. Similarly, taking u positive and very close to 0 , we obtain that $4a+b$ must be nonnegative. Thus $4a+b=0$ and so the sum of roots of P is $-\frac{b}{a}=4\text{.}$