<s>[INST] Find $x$ such that $\left\vert -a x\pm \sqrt{a^2x^2-1}\right\vert\leq 1.$ Given inequality
\begin{align*}
 \left\vert -a x\pm \sqrt{a^2x^2-1}\right\vert\leq 1.
\end{align*}
for
$-1\leq a\leq 1,\quad a^2x^2-1\geq 0,\quad$ $x\in\mathbb R$.
Determine the what the range of $x$?.
I try as follows.
\begin{align}
&\left\vert -a x\pm \sqrt{a^2x^2-1}\right\vert\leq 1\\
\iff& \left\vert -a x\right\vert+\left\vert\pm \sqrt{a^2x^2-1}\right\vert\leq 1 \quad (\text{triangle inequality})\\
\iff& \left\vert a\right\vert  \left\vert x\right\vert+\sqrt{a^2x^2-1}\leq 1 \quad (\text{absolute value properties})\\
\iff&  \left\vert x\right\vert+\sqrt{x^2-1}\leq 1 \quad (\max a =1)\\
\iff&  \left\vert x\right\vert+\vert x\vert\leq 1 \quad (\sqrt{x^2-1} \leq \sqrt{x^2}=\vert x\vert)\\
\iff&  \vert x\vert\leq \frac{1}{2}\\ 
\iff&  -\frac{1}{2}\leq x\leq \frac{1}{2}.\\ 
\end{align}
Now take $a=0.5$ and $x=2$. Then
$$\vert-0.5\cdot 2\pm \sqrt{0} \vert= 1\leq 1.$$
Why if I take $a=0.5$ and $x=2$ then inequality satisfied? Even though I have found the range of $x$ in $-\frac{1}{2}\leq x\leq \frac{1}{2}$.
I feel in finding $x$ is correct, but the result is incorrect. Anyone know my mistake?
 [/INST] The triangle inequality won't help because you want transformations that are equivalences.
Setting $y=ax$ and squaring both sides:
$$y^2 \mp 2y \sqrt{y^2-1} +y^2-1\leqslant 1$$
then rearranging gives
$$ y^2 - 1 \leqslant \pm y \sqrt{y^2-1}$$
which holds if $y^2=1$, or if $y^2>1$ we divide by the square root to get
$$\sqrt{y^2-1} \leqslant \pm y$$

*

*If the inequality is supposed to hold in either $\pm$ cases, then it follows that $y=0$ and the square root must be zero, too. However, the root is only defined if $|y|\geqslant 1$.


*If just one of the $\pm$ cases must hold, then we may use $$\sqrt{y^2-1}\leqslant |y|$$ which always holds true.
</s>