How can I solve this inequality? Have a nice day, how can I solve this inequality?
$$a<b<-1$$
$$ |ax - b| \le  |bx-a|$$
what is the solution set for this inequality
 A: Squaring both sides,
$$a^2x^2-2abx+b^2 \leq b^2x^2-2abx+a^2$$
$$(a^2-b^2)x^2 \leq a^2-b^2$$
Since $a^2-b^2>0$, we can divide both sides by $a^2-b^2$ and preserve the sign of inequality.
$$x^2 \leq 1$$
$$|x|\leq 1$$
$$-1 \leq x \leq 1$$
A: There may be more efficient ways to do these but for clarity I like to break absolute values into cases:
Case 1: $ax -b \ge 0$ and $bx -a \ge 0$.
[This implies $ax \ge b\implies x \le b/a$ and likewise $x \le a/b$ so $x \le \min(a/b,b/a) = b/a < 1$.  Let's keep in mind $b/a < 1 < a/b$]
Then $|ax - b| \le |bx -a| \implies ax - b \le bx - a$
So $(a-b)x \le (b-a)$; $a < b$ so $(a - b) < 0$ so
$x \ge (b-a)/(a-b) = -1$
So $-1 \le x \le b/a < 1$.  OR
Case 2:  $ax -b <0$ and $bx-a < 0$.
[This implies $ax < b\implies x > b/a$ and $x > a/b$ so $x > \max(a/b,b/a) =a/b > 1$]
then $b - ax \le a - bx$ so
$(b - a) \le (a - b)x$ so
$(b-a)/(a-b) \ge x$ so $x \le -1$ which is a contradiction.
OR
Case 3:  $ax -b < 0$ and $bx - a \ge 0$
[which implies $ax < b\implies x >b/a > 1$ and $bx \ge a\implies x \le a/b \le 1$.  So $b/a < x \le a/b$.  Remember $b/a < 1 < a/b$]
So $b - ax \le bx - a$ so $(b+a) \le (a + b)x$.  Now (a+b) < 0. So
$(b+a)/(a+b) = 1 \ge x$
So $b/a < x \le 1$.
OR
Case 4:  $ax - b \ge 0$ and $bx - a < 0$
[which implies $ax \ge b \implies x\le b/a$ and $x > a/b$ so $a/b < x \le b/a$ which is a contradiction.]
So we know $-1 \le x \le b/a$ OR $b/a < x \le -1$ so combining the two results we know $-1 \le x \le 1$.
But there are probably more efficient ways to do this.
