<s>[INST] 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...
 [/INST] You made a mistake. From here
$$ a^2 - (1 + b)a + b \geq 0$$
you went here
$$ a=\frac{-(1 + b) + \sqrt{4 + (1 + b)^2}}{2} $$
While the correct solution is
$$a=\frac{-(1 + b) + \sqrt{ (1 + b)^2-4b}}{2} =\frac{-(1 + b) + \sqrt{ (1 - b)^2}}{2}$$
$$a_1=1;\;a_2=b$$
$$a\le 1\lor a\ge b$$
that is
$$x^2\le 1 \lor x^2\ge (m+1)^2$$
or
$$-1\le x \le 1 \lor  x\ge| m+1|$$
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