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I'm a little confusing in proving this inequality $$\frac{a+b}{|c-b|}<1$$ where $a,b,c$ are positive real numbers, and $a<c$. any help!

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$a=4$, $b=3$, $c=4$ –  N. S. Nov 17 '11 at 13:46
    
In the comment above (probably made before an editing) $a$ is not smaller than $c$. But anyway you need something that doesn´t let $|c-b|$ to be arbitrarily small. –  Giovanni De Gaetano Nov 17 '11 at 14:00
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N.S. obviously meant $a=3$, $b=3$, $c=4$ :) (So, the inequality is false.) –  David Mitra Nov 17 '11 at 14:05
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How about $ a=1, b=1, c=2$? –  ofer Nov 17 '11 at 14:07
    
I think the inequalities may be the wrong way round. Reverse both and you have something which works. –  Mark Bennet Nov 17 '11 at 16:00

1 Answer 1

The inequality is wrong. Choose $a=b=1$ and $c=2$ then $a<c$ but

$$\frac{a+b}{|c-b|}=\frac{2}{1}=2$$

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