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 $P(x).$
Let $P(x) = ax^2 + bx + c.$  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
\[ax^6 + ax^4 + bx^3 - (a + b)x^2 + bx - a - b \ge 0.\]This factors as
\[(x - 1)(x^2 + 1)(ax^3 + ax^2 + ax + a + b) \ge 0.\]For this inequality to hold for all real numbers $x,$ $ax^3 + ax^2 + ax + a + b$ must have a factor of $x - 1.$  (Otherwise, as $x$ increases from just below 1 to just above 1, $x - 1$ changes sign, but $(x^2 + 1)(ax^3 + ax^2 + ax + a + b)$ 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 $ax^2 + bx + c = 0$ is $-\frac{b}{a} = \boxed{4}.$