Let $P(x) = ax^2+bx+c$. Then $\Delta(x) = P(x^3+x) - P(x^2+1) =$ $ P(x(x^2+1)) - P(x^2+1) =$ $ (x-1)(x^2+1)(a(x+1)(x^2+1) + b)$. Since we need $\Delta(x) \geq 0$ for all $x$, we need $x-1 \mid a(x+1)(x^2+1) + b$, i.e. $4a + b = 0$. Therefore $b= -4a$, and so by Viète's relation, the sum of the roots of $P(x)$ is $\boxed{x_1 + x_2 = -\dfrac {b} {a} = 4}$. Indeed $\Delta(x) = a(x-1)^2(x^2+1)(x^2+2x +3) \geq 0$ for all $x$ (since the discriminant of $x^2+2x +3$ is negative), provided we take $a>0$. Therefore all such polynomials are of the form $P(x) = k^2(x^2 - 4x + C)$.

mikolez wrote: Let $P(x)$ be a quadratic trinomial, so that for all $x\in \mathbb{R}$ the inequality $P(x^3+x)\geq P(x^2+1)$ holds. Find the sum of the roots of $P(x)$. Another method (a bit too late) Writing $f(x)=P(x^3+x)-P(x^2+1)$, we get $f(x)\ge 0$, $C_{\infty}$ and $f(1)=0$ and so $f'(1)=0$ and so $P'(2)=0$ Writing $P(x)=ax^2+bx+c$ with $a\ne 0$, then $P'(2)=0$ becomes $4a+b=0$ and so $-\frac ba=4$ And so, if such a quadratic exists the sum of roots is $\boxed{4}$ Btw, such a polynomial exists. For example $P(x)=x^2-4x$

Heres another method (bit too late ): Note that the leading co-efficient of $P(x)$ is positive. If its negative, then for large enough values of $x$, $P(x)$ is decreasing, but as $x^3+x > x^2+1$ this implies $P(x^3+x) < P(x^2+1)$, contradiction. Let $P(c)$ be the minimum value of $P(x)$ over all real. Note that, $x^3+x=c$ (for only one real value of $x$) implies $P(c) = P(x^3+x) \ge P(x^2+1) \ge P(c)$. Thus: $x^3+x=x^2+1$ which implies $x=1$ and $c=2$. Therefore, sum of roots is $2c=\boxed{4}$.