Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$ Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$.
I tried using substitution as follows:
$$ a = x^2 $$
$$ b = (m + 1)^2$$
Using the substitution:
$$ a^2 - (1 + b)a + b \geq 0$$
Substituting in the second degree equation:
$$ \frac{-(1 + b) + \sqrt{4 + (1 + b)^2}}{2} $$
Solving the binomial coefficients $(m + 1)^4$ and $(m + 1)^2$, I got:
$$ \frac{-(1 + (m + 1)^2) + \sqrt{m^4 + 4m^3 + 8m^2 + 8m + 8}}{2} $$
From here I don't know how to proceed...

Your method might work, but I couldn't really see how to proceed. A substitution that I am sure would work is as follows:
First expand your polynomial: $x^4-(m^2+2m+2)x^2+m^2+2m+1$.
Next, you can use your $a=x^2$ substitution: $a^2-(m^2+2m+2)a+m^2+2m+1$. Then, you use essentially the same $b$ substitution, but expanded: $b=m^2+2m+1$. Once you do this you get $a^2-(b+1)a+b$, you solve and get $a=1, a=b$ (verify yourself that they work). Then substitute back $a$ and $b$: $x^2=1, x^2=m^2+2m+1$, so your solutions for $x$ are: $x=1$, $x=-1$, $x=m+1$, $x=-m-1$. Therefore, since an extremely large or small value of $x$ will make the equation positive, your range of x-values are $(-\infty, -|m|]\cap[-1, 1]\cap[|m|, \infty)$. Note when $m=-2$, your range of x-values are $(-\infty, \infty)$
Edit: My steps are essentially your steps (expanded out), but I didn't use quadratic formula like on the fourth step, but factored it.
Using what you have:
$ a = x^2 $
$ b = (m + 1)^2$
$ a^2 - (1 + b)a + b \geq 0$
$a=b, a=1$ are solutions to the inequality, and solve for bounds
The rest is the same.
