Prove Inequality: |1-K+x|/|1+y| < 1

[Only a Mirage]

Homework Statement

Given:

|x-y| &lt; K

x+y &gt; K - 2

0 &lt; K &lt; 1

Prove:

\frac{|1-K+x|}{|1+y|} &lt; 1

The Attempt at a Solution

I have tried using the fact that |x-y| &lt; K \Rightarrow -K &lt; x-y &lt; K \Rightarrow y-K &lt; x &lt; y+K to write \frac{1-K+x}{1+y} &lt; \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.

 

[jbunniii]

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

 

[Only a Mirage]

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

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.

 

[jbunniii]

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

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 $y + 1 > 0$.

 

[Only a Mirage]

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 $y + 1 > 0$.

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