if the polynomial $x^4+ax^3+2x^2+bx+1=0$ has four real roots ,then $a^2+b^2\ge 32?$ if such that the polynomial
$$P(x)=x^4+ax^3+2x^2+bx+1=0$$
has four  real roots. prove or disprove  $$a^2+b^2\ge 32?$$
I have solve this problem: if the polynomial $P(x)$ at least have one real root,then $a^2+b^2=8$,see a^2+b^2\ge 8,and when $P(x)=x^4+2x^3+2x^2+2x+1=(x+1)^2(x^2+1)$ but if the $P(x)$ have four real roots,I can't it.

We will show that with some symmetry consideration we can either reduce to the case where all the roots have the same sign and then the result follows since if say $x_k >0$, $\sum {x_k} =|a| \ge 4(x_1..x_4)^{\frac{1}{4}} =4, a^2 \ge 16$ and then $\sum {1/x_k}=|b|$ and same applies so $b^2 \ge 16$ and we are done, or we must anyway have $|a|, |b| \ge 4$ and again the result follows.
Replacing $x \to 1/x$ changes $a$ into $b$ keeping the roots real so we can assume wlog $|a| \ge |b|$; replacing $x \to -x$ changes the signs of $a,b$ so we can assume wlog $a \ge 0$. But then if $x \ge 1, ax^3 \ge \pm bx$ since $a \ge |b|$ so $x^4+ax^3+2x^2+bx+1 >0$, while if $0 \le x \le 1, b=-c, a \ge c>0$ and we get the equation:
$x^4+ax^3+2x^2-cx+1=0, 0<c\le a$ and there is at least a root $0<x_0<1$ (sum of roots is negative so some are negative, product is $1$ so this means two are neagtive, two positive and our choices imply that the posiitive ones are in $(0,1)$)
Noting that the equation can be rewritten as:
$(x^2+2x-1)^2+(a-4)x^3+(4-c)x=0$ or $(x^2+2x-1)^2 =(c-4)x+(4-a)x^3$ we immediately get that if $c<4$ we must have $a<4$ too but then $4-a \le 4-c$ since $c \le a$ hence remembering that we have a root $0<x_0<1$ we get $0 \le (x_0^2+2x_0-1)^2 =(4-a)x_0^3-(4-c)x_0 < (4-a)x_0-(4-c)x_0 \le 0$ so a contradiction! Hence $|b|=c \ge 4, a \ge c \ge 4$ and we are done!
