Inequality |y|>|y+1|

ABSOLUTE VALUE PROPERTIES:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

BACK TO THE QUESTION:

\(|y|>|y+1|\)

APPROACH #1:

\(|y|>|y+1|\) --> we have two transition points for 0 and -1 (transition point is the value of y for which an expression in the modulus equals to zero). Thus we have three ranges to check:

1. \(y<-1\) --> in this case \(|y|=-y\) and \(|y+1|=-(y+1)\), so we'd have \(-y>-(y+1)\) --> \(0>-1\) --> TRUE. So, when \(y<-1\) inequality \(|y|>|y+1|\) holds true.

2. \(-1\leq{y}\leq{0}\) --> in this case \(|y|=-y\) and \(|y+1|=y+1\), so we'd have \(-y>y+1\) --> \(y<-\frac{1}{2}\). So, when \(-1\leq{y}<-\frac{1}{2}\) inequality \(|y|>|y+1|\) holds true.

3. \(y>0\) --> in this case \(|y|=y\) and \(|y+1|=y+1\), so we'd have \(y>y+1\) --> \(0>1\) --> FALSE. So, when \(y>0\) inequality \(|y|>|y+1|\) has no solution.

So, we have that \(|y|>|y+1|\) holds true for \(y<-\frac{1}{2}\) (combine true ranges from 1 and 2).

APPROACH #2:

Since both parts of the inequality are non-negative, then we can square it: \(y^2>y^2+2y+1\) --> \(y<-\frac{1}{2}\).




Hope this helps.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Thank you.

Thank you for the awesome explanation!