Find the minimum possible value of $a^2+b^2$ where $a$, $b$ are two real numbers such that the polynomial $$x^4+ax^3+bx^2+ax+1,$$ has at least one real root.
My attempt: Let p be a real root. Therefore $p^4 + a(p^3) + b(p^2) + ap + 1 = 0$.
Divide both sides by $p^2$, noting that $p$ cannot be zero since $P(0) = 1$, we get $$p^2 + ap + b + a/p + 1/(p^2) = 0.$$
I rearranged this into a quadratic; i.e. $$(p+1/p)^2 + a(p+1/p) + (b-2) = 0.$$
If $p$ is real, $p+1/p$ is real, so discriminant is non-negative. setting discriminant $\geq0$, I get:
$$a^2 - 4(b-2)\geq0,$$
$$a^2 \geq 4b-8,$$
$$a^2 + b^2 \geq b^2 + 4b - 8 = (b+2)^2 - 12 \geq -12$$ But this is useless because it is obvious $a^2 + b^2 \geq 0$.
I believe this is because I also need to use the fact that $|p+1/p| \geq 2$, however I am unsure how to use this inequality with the discriminant.
Thanks in advance