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.

Edit : I should have added a condition that $|-a/2|\ge 2$. 

I think that your idea is nice, but note that $(2)$ is incorrect. Using $x+\frac 1x=y$, we have
$$y^2+ay+b\color{red}{-2}=0\tag3$$
So, we have to have
$$a^2-4(b-2)\ge 0\iff b\le \frac{a^2}{4}+2\tag4$$
We want $(3)$ to have at least one real solution $y$ such that $|y|\ge 2$ as you wrote.
Let $f(y):=y^2+ay+b-2$. The condition is represented as
$$\left|-\frac a2\right|\ge 2\quad\text{or}\quad f(-2)\le 0\quad\text{or}\quad f(2)\le 0$$$$\iff |a|\ge 4\quad\text{or}\quad 2-2a+b\le 0\quad\text{or}\quad 2+2a+b\le 0\tag5$$
Hence, drawing $(4)(5)$ in $ab$-plane gives that the minimum value of $a^2+b^2$ is
$$\left(\frac{|2\pm 2\cdot 0+0|}{\sqrt{(\pm 2)^2+1^2}}\right)^2=\color{red}{\frac 45}$$
(note here that $a^2+b^2$ represents the square of the distance from $(0,0)$ to $(a,b)$)
for $(a,b)$ such that $a^2+b^2=\frac 45$ and $b=\pm 2a-2$, i.e. $(a,b)=(\pm 4/5,-2/5)$.
