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...

Here's how it's done in high school, using Vieta's relations and results on the sign of a quadratic polynomial:
First set $y=x^2$ and solve for $y$: you have the quadratic inequation
$$y^2-\bigl(1+(m+1)^2\bigr)y+(m+1)^2 \ge0,\qquad y\ge 0.$$
The discriminant is
$$\Delta=\bigl(1+(m+1)^2\bigr)^2-4(m+1)^2=\bigl(1-(m+1)^2\bigr)^2,$$
therefore the quadratic polynomial in $y$ has two roots $y_0$ and $y_1\;(y_0\le y_1)$.
Furthermore these roots are nonnegative, since $y_0+y_1=\bigl(1+(m+1)^2\bigr)^2>0\:$ and $\:y_0y_1=(+1)^2\ge 0$.
Conclusion; the original inequation is equivalent  to
$$0\le y\le y_0\;\text{ or }\;y\ge y_1\iff \begin{cases}-\sqrt{y_0\mathstrut}\le x\le \sqrt{y_0\mathstrut},\\
\quad\text{ or }\\ x\le -\sqrt{y_1\mathstrut},\enspace x\ge\sqrt{y_1\mathstrut}. \end{cases}$$
