Condition for quartic polynomial coefficients given at least one real root 
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
 A: Hint:
Put $t=x+\frac{1}{x}$ then: 
$x^4+ax^3+bx^2+ax+1=0$ has a real root if and only if $t^2+at+(b-2)=0$ has a real root $t$ such that $|t| \ge 2$, which is equivalent to 
\begin{equation}\tag{*}
\left[\begin{matrix}
\frac{-a-\sqrt{a^2-4(b-2)}}{2} \le -2 \\ 
\frac{-a+\sqrt{a^2-4(b-2)}}{2} \ge 2
\end{matrix}\right.
\end{equation}
Using this:
$$A \le \sqrt{B} \Longleftrightarrow
\left[\begin{matrix}
\left\{\begin{matrix}
B \ge 0 \\ 
A < 0
\end{matrix}\right. \\
\left\{\begin{matrix}
A \ge 0 \\ 
A^2 \le B
\end{matrix}\right. 
\end{matrix}\right.
 $$
After some operations we have $(*)$ is equivalent to
$$\left[\begin{matrix}
(1)\left\{\begin{matrix}
a \ge 4 \\ 
a^2-4(b-2) \ge 0
\end{matrix}\right. \\
(2)\left\{\begin{matrix}
a \le -4 \\ 
a^2-4(b-2) \ge 0
\end{matrix}\right. \\
(3)\left\{\begin{matrix}
a < 4 \\ 
2a -b \ge 2
\end{matrix}\right. \\
(4)\left\{\begin{matrix}
a > -4 \\ 
2a+b \le -2
\end{matrix}\right. 
\end{matrix}\right.$$
For $(1)$ and $(2)$: $a^2+b^2 \ge 4^2+0 =16$. 
For $(3)$: using $5(a^2+b^2) = (2a-b)^2+(a+2b)^2$ we have $a^2+b^2 \ge \frac{4}{5}$, attained when $a=\frac{4}{5},b=-\frac{2}{5}$.
For $(4)$: using $5(a^2+b^2) = (2a+b)^2+(a-2b)^2$ we have $a^2+b^2 \ge \frac{4}{5}$, attained when $a=-\frac{4}{5},b=-\frac{2}{5}$.
Conclusion: the minimum value of $a^2+b^2$ is $\frac{4}{5}$, attained when $(a,b)=\left(\pm\frac{4}{5},-\frac{2}{5}\right)$.
A: Replacing $a$ with $-a$ only changes the signs of the zeros, so w.l.o.g. we can assume that $a\ge0$. The quadratic with the unknown $p+1/p$ (nice trick, BTW!) that you derived gives
$$
p+\frac1p=\frac{-a\pm\sqrt{a^2-4(b-2)}}2.
$$
Given the assumption $a\ge0$ we see that of these two alternatives the solution with a minus sign gives the larger value to $|p+1/p|$. Therefore the condition $|p+1/p|\ge2$, now rewritten as $p+1/p\le-2$, gives us another constraint
$$
-a-\sqrt{a^2-4(b-2)}\le-4.
$$
This is equivalent to
$$
4-a\le\sqrt{a^2-4(b-2)}.\qquad(*)
$$
We shall see that there are solutions with $a^2+b^2<16$, so w.l.o.g. we can further assume that $a<4$, so $4-a>0$ and we can square both sides of $(*)$ arriving at
$$
16-8a+a^2\le a^2-4b+8\implies a\ge (b+2)/2.
$$
Again, later developments will reveal that $b+2$ must be positive at the sought minimium, so to minimize $a^2$ we must have equality here, i.e. $a=(b+2)/2$. Thus we really want to minimize
$$
a^2+b^2=\frac14[(b+2)^2+4b^2]=\frac14[5b^2+4b+4].
$$
It is trivial to show that this has a minimum at $b=-2/5$. The corresponding
$a=(b+2)/2=4/5$, and at this point we have
$$
a^2+b^2=\frac{4^2+2^2}{25}=\frac45.
$$
If either the assumption $4-a\ge0$ or the assumption $b+2\ge0$ (that I made while looking for this candidate point) were invalid, then clearly $a^2+b^2$ would have a larger value, so we can dismiss those possibilities.
As a final check we see that when $a=4/5, b=-2/5$ your polynomial has a double root at $x=-1$, not unexpectedly matching with $p+1/p=-2$.
