- If |x| ≥ |y|, then “x ≥ y” OR “x ≤ -y”
- If |x| ≤ |y|, then “x ≤ y” AND “x ≥ -y”
Are these rules true? Are there any other factors I should take care of while evaluating such inequalities? (such as |x| ≥ 0 or something)
Yeah, it looks like it holds iff y \geq 0, but you can do better to bound x for all values of y regardless of its sign.
If I had to find the range of x and y values for |x + 1| ≥ |y + 1|
Is this way correct?
- If x + 1 ≥ 0 [i.e x ≥ -1]
- x + 1 ≥ y + 1
- x ≥ y
- If x + 1 ≤ 0 [i.e x ≤ -1]
- x + 1 ≤ - y - 1
- x ≤ -y - 2
Final Answer:
x ≥ y and x ≥ -1 OR x ≤ -y - 2 and x ≤ -1
But you just treated |y + 1| as y + 1, which is only true for y such that y \geq -1.
I would suggest squaring both sides instead:
\begin{aligned}
|x + 1|^2 \ge |y + 1|^2 &\iff (x + 1)^2 \ge (y + 1)^2 \\
&\iff (x - y)(x + y + 2) \ge 0
\end{aligned}
Perhaps you can take it from here.
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Would we be able to use it for such inequalities as well |x + 1| ≥ |y + 1| - 2?
- We can probably just add “2” on both sides, square both sides, and then take it from there? The calculation becomes a bit complex though
You wouldn’t be using the “squaring” method for this because you won’t get rid of all instances of absolute values.
Instead, you could decompose your inequality into intervals:
-
|x + 1| = x + 1 when x \ge -1 and -(x + 1) otherwise
-
|y + 1| = y + 1 when y \ge -1 and -(y + 1) otherwise
In total, you have 4 distinct cases to consider, each of which implies a different inequality.
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