jmc

Let $P(x)=a{x}^2+bx+c.$ Then $$a(x^3+x{)}^2+b(x^3+x)+c\ge a(x^2+1{)}^2+b(x^2+1)+c$$for all real numbers $x.$ This simplifies to $$a{x}^6+a{x}^4+b{x}^3-(a+b)x^2+bx-a-b\ge 0.$$This factors as $$(x-1)(x^2+1)(a{x}^3+a{x}^2+ax+a+b)\ge 0.$$For this inequality to hold for all real numbers $x,$ $a{x}^3+a{x}^2+ax+a+b$ must have a factor of $x-1.$ (Otherwise, as $x$ increases from just below 1 to just above 1, $x-1$ changes sign, but $(x^2+1)(a{x}^3+a{x}^2+ax+a+b)$ does not, meaning that it cannot be nonnegative for all real numbers $x.$) Hence, setting $x=1,$ we get $a+a+a+a+b=0,$ so $4a+b=0.$

Then by Vieta's formulas, the sum of the roots of $a{x}^2+bx+c=0$ is $-\frac{b}{a}=\boxed{4}.$