# Inequality problem

1. Aug 1, 2013

### Only a Mirage

1. The problem statement, all variables and given/known data

Given:

$|x-y| < K$

$x+y > K - 2$

$0 < K < 1$

Prove:

$\frac{|1-K+x|}{|1+y|} < 1$

3. The attempt at a solution

I have tried using the fact that $|x-y| < K \Rightarrow -K < x-y < K \Rightarrow y-K < x < y+K$ to write $\frac{1-K+x}{1+y} < \frac{1+y}{1+y} = 1$

But I can't figure out how to show that the absolute value is less than one.

I have also been trying various applications of the triangle inequality with little success.

Any help would be greatly appreciated.

Last edited: Aug 1, 2013
2. Aug 1, 2013

### jbunniii

Hint: what does the given inequality $x + y > K - 2$ tell you about $1 - K + x$?

3. Aug 2, 2013

### Only a Mirage

It tells me that $1 - K + x > -(y+1)$

which I actually had written on my paper before, but thanks to you I think I've realized the connection...

Since I already had from the first assumption $1-K+x<1+y$, I now have

$-(y+1)<(1-K+x)<(y+1)$, which seems to imply $|1-K+x| < (y+1)$ as desired.

Since we didn't know before that $|y+1| = y+1$, it seems that this claim actually follows from the assumptions? I think it does since I think

$-A<T<A \Rightarrow A>0$, if both inequalities are satisfied, right?

Thank you so much for your help.

4. Aug 2, 2013

### jbunniii

Yes, this implication is correct. Ignoring the $T$ in the middle of the inequality chain, you have $-A < A$, so $2A > 0$, hence $A > 0$.

Note that the fact that $y + 1 > 0$ is important, because you are dividing both sides of each inequality by this expression. If $y+1$ was zero, the division would be undefined, and if $y+1$ was negative, the direction of the inequality would flip. Fortunately both of these possibilities are excluded by $y + 1 > 0$.

5. Aug 2, 2013

### Only a Mirage

That makes sense to me. Thanks again for the help.