<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] Following the steps given in Wikipedia, we can set the problem as
$$\text{Minimize}[a^2+b^2,\{a,b\}]$$
subject to the three inequality constraints
$$P=16-3a^2 <0$$
$$D=-3 a^4+32 a^2-16 a b \leq 0$$
$$\Delta=-27 a^4-4 a^3 b^3+36 a^3 b-2 a^2 b^2+256 a^2+36 a b^3-512 a b-27 b^4+256 b^2 \leq 0$$
This is a quite simple task, leading to the solutions $a=\pm4$ and $b=\mp4$ and then $a^2+b^2 \geq 32$.
The symmetry, already underlined in other answers is obvious looking at the constraints.
All the above was done using calculus. It could be done without any if, as in many optimization problems of the kind, we assume that the solution will correspond to the saturation of inequality constraints. If this is the case, then
$$D=0 \implies b=2 a\left(1-\frac{3 }{16}a^2\right)$$
$$\Delta=0 \implies a^2(a^2-16)^2(459 a^6-7776 a^4+43776 a^2-65536)=0$$
$a^2=0$ must be rejected since $P<0$ would not be satisfied.
Concerning the cubic in $a^2$, it shows only one real root which is
$$a^2=\frac{32}{51} \left(9-\cosh \left(\frac{1}{3} \cosh ^{-1}(577)\right)\right) \approx 2.32647$$to be rejected since $P<0$ would not be satisfied $(2.32647 < \frac {16}3)$.
All of that makes that the only solution is $a^2=16$ to which corresponds $b^2=16$ too.
One thing I did not have time to do is to look at the shadow prices of this problem.
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