<s>[INST] 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.
 [/INST] 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!
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