Find the minimum value of $a^2+b^2$ 
Let a and b be real numbers for which the equation $x^4 +ax^3 +bx^2
 +ax+1=0 \tag1$ has at least one real solution. For all such pairs $(a,b)$, find the minimum value of $a^2 + b^2$.


Using $x + \frac 1 x = y$ in (1):
$y^2 + ay+b-2=0 \tag2$ therefore the first condition is $a^2 - 4b + 8\ge 0$.
The second one, coming from $x^2 -yx + 1=0$, is $y^2 - 4 \ge0$.
The calculus is a mess, so I don't think this is the way to solve it. Does anyone have a smarter idea?
UPDATE
I've corrected (2) following @mathlove suggestion.

Divide the equation by $x^2$
$$
\begin{align}
0
&=x^2+ax+b+\frac ax+\frac1{x^2}\\
&=\left(x+\frac1x\right)^2+a\left(x+\frac1x\right)+(b-2)\tag{1}
\end{align}
$$
Since $\left|\,x+\frac1x\,\right|\ge2$ and the solutions to $(1)$ are
$$
x+\frac1x=\frac{-a\pm\sqrt{a^2-4b+8}}2\tag{2}
$$
and since the sign of $a$ does not change $a^2+b^2$, and only changes the sign of $x$, we can assume that $a\ge0$. Then we need
$$
a+\sqrt{a^2-4b+8}\ge4\tag{3}
$$
Since $a=1$ and $b=0$ satisfy $(3)$, the minimum of $a^2+b^2$ is at most $1$. Therefore, we can assume that $a\le1$ and $b\ge-1$. Then $(3)$ becomes
$$
a^2-4b+8\ge16-8a+a^2\iff a\ge\frac{b+2}2\tag{4}
$$
Therefore,
$$
a^2+b^2\ge\left(\frac{b+2}2\right)^2+b^2\ge\frac45\tag{5}
$$
Since $a=\frac45$ and $b=-\frac25$ satisfies $(3)$ and $a^2+b^2=\frac45$, we get
$$
\min\!\left(a^2+b^2\right)=\frac45\tag{6}
$$
