Let $P$ be a polynomial, it has $n$ distinct real roots and all these roots are larger than 1. Let $$ Q(x)=(x^2+1)P(x)P'(x)+x[(P(x))^2+(P'(x))^2]. $$
Prove $Q(x)$ has at least $2n-1$ distinct real roots.
I start to find $Q(x)=0$ between two roots. Assume $P(x)<0$ in $(x_1,x_2)$ and $P(x_1)=P(x_2)=0$. There exists a minimal value at $\xi\in(x_1,x_2)$ s.t. $$ P(\xi)<0, P'(\xi)=0, P''(\xi)>0. $$ Then I get $P(\xi)+P'(\xi)<0$.
It can be deduced by contradiction that not all $x\in (x_1,x_2)$ satisfy $P(x)+P'(x)\leq 0$. Hence there exists $\eta$ s.t. $P(\eta)+P'(\eta)>0$, then there exists $\alpha$ s.t. $$P(\alpha)+P'(\alpha)=0.$$ $Q(\alpha)<2xP(\alpha)P'(\alpha)+x[((P(\alpha))^2+(P'(\alpha))^2]\leq 0$, i.e. $$Q(\alpha)<0.$$ If I have $Q(x_1)>0$ and $Q(x_2)>0$ then by mean value theorem there are two roots in $(x_1,x_2)$.
The problems of this thought are:
(1) Even I have two roots in every interval, only $2n-2$ roots be found;
(2) $Q(x_1)>0$ and $Q(x_2)>0$ may be zero and my process would be useless?
Hope someone give me an answer or a hint.