# Thread:

1. ## inequality problem

The real number c is positive and the real numbers a and b are such that $ab > 0$ and $a > b$. How can I disprove the case $a - c < b + c$ for all such a, b, and c? (The case $\frac{c}{a} < \frac{c}{b}$ works). When I tried it, I can't seem to get numbers that disprove it??

2. Originally Posted by donnagirl
The real number c is positive and the real numbers a and b are such that $ab > 0$ and $a > b$. How can I disprove the case $a - c < b + c$ for all such a, b, and c? (The case $\frac{c}{a} < \frac{c}{b}$ works). When I tried it, I can't seem to get numbers that disprove it??
How hard did you try? How about a= 100, b= 1, c= 1?

3. Hello donnagirl
Originally Posted by donnagirl
The real number c is positive and the real numbers a and b are such that $ab > 0$ and $a > b$. How can I disprove the case $a - c < b + c$ for all such a, b, and c? (The case $\frac{c}{a} < \frac{c}{b}$ works). When I tried it, I can't seem to get numbers that disprove it??
I'm not quite sure I understand the problem here, because it seems very easy to find a counter-example. In order that $a-c>b+c$, we must ensure that $a-b > 2c$. So, for instance:
$a=10, b = 1, c = 3$
gives $a-c=7$ and $b+c = 4$. So $a-c> b+c$.

Grandad

4. Thanks again Grandad!
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