Quadratic polynomial

I posted the solution to this question on Art of problem solving a while ago, so I am going to copy paste it instead of typing again.

Solution :
Let $P(r) = ar^{2} + br + c \rightarrow$ Sum of roots $= \dfrac{-b}{a}$
For simplicity, $x^{3} + x = t$ & $x^{2} + 1 = q$
$\therefore P(t) \ge P(q)$
$\therefore at^{2} + bt + c \ge aq^{2} + bq + c$
$a(t^{2}-r^{2}) + b(t-r) \ge 0$
$(t-r)(a(t+r) + b \ge 0$
$\left(x^{3} - x^{2} +x-1\right)\left(a(x^{3}+x^{2}+x+1) + b\right) \ge 0$ $x \in R$
$(x^{2}+1)(x-1)(a(x^{2}+1)(x+1) + b) \ge 0$ for all $x \in R$
$\therefore (x-1)(a(x^{2}+1)(x+1) + b) \ge 0$ for all $x \in R$

For x > 1,
$a(x^{2}+1)(x+1) + b \ge 0$
For x < 1 $a(x^{2}+1)(x+1) + b \le 0$
Thus, equality holds for one, i.e $x = 1$ is a root of $(a(x^{2}+1)(x+1) + b)$
$\therefore a(2)(2) + b = 0$
$\therefore \dfrac{-b}{a} = 4$
Thus, the sum of roots of $P(r)$ is $4$

Let $P(x) = ax^{2} + bx + c$ , and $y = P(x^{3} + x) - P(x^{2} + 1)$ .

We need to find the value of $\frac{-b}{a}$ .

It is easy to get $y = P(x^{3} + x) - P(x^{2} + 1) = a(x^{2} + 1)^{2}(x^{2} -1) + b(x^{2}+1)(x-1) \geq 0$ .

Let $b=ka$ for some real number $k$ . Therefore, $y= a(x^{2} + 1)^{2}(x^{2} -1) + ka(x^{2}+1)(x-1)= a(x^{2}+1)(x-1)[(x^{2}+1)(x+1) + k]$ .

WLOG $a \geq 0$ . We got $y= a(x^{2}+1)(x-1)[(x^{2}+1)(x+1) + k]\geq (x-1)[(x^{2}+1)(x+1) + k]$ .

Let $Q(x) = (x^{2}+1)(x+1) + k = R(x) + k$

(i) First Case : $x = 0$ , $y= 0$

(ii) Second Case : $x > 1$ , $y\geq Q(x)$ .

(iii) Third Case : When $x < 1$ , $Q(x)$ has to be negative in order for the original inequality to be satisfied.

It is obvious that $Q(1) = R(1) + k = 4 + k$ and $R(x)$ is a monotonically increasing function. Therefore $R(m) \leq 4 \forall m \leq 1$ . Thus, $Q(x) = R(x) - 4 < 0 \forall x < 1$ , satisfying the third case.

Again, $R(x)$ is monotonically increasing, therefore $R(m) > 4 \forall m > 1$ . Therefore $Q(x) > 0 \forall x > 1$ . Satisfying the second case. It implies that $k= -4$ satisfies all the case.

Thus, $b= ka= -4a$ , implying $\frac{-b}{a} = 4$ .

Thus the sum of the roots of $P(x)$ is equal to 4.

Here is a interesting way to look towards the problem, x^2+1 > 1 1. Let x>1 x^2+1>2 x(x^2 + 1) > x^2 + 1 > 2 2. Let 0<x<1 1< x^2 + 1< 2 0<x(x^2+1)<x^2+1<2 therefore at 2 the minima of parabola occur hence sum of roots/2 = 2