Proving ab - a - b + 2 > 1 in Simple Algebra: Tips and Hints

Thread starter Chen
Start date Mar 28, 2005
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Algebra Proof

AI Thread Summary

To prove that ab - a - b + 2 > 1 for a > 1 and b > 1, it can be rewritten as ab - a - b + 1 > 0. This simplifies to a(b - 1) - b + 1 > 0, which can be rearranged to -a(b - 1) + b - 1 < 0. The final form shows that (b - 1)(1 - a) < 0, indicating the relationship between a and b. This transformation clarifies the proof and highlights the conditions under which the inequality holds true.

Mar 28, 2005

Chen

Hi,

I need to prove that for every a > 1 and b > 1 the following holds:

ab - a - b + 2 > 1

Can someone please throw me some hints? :)

Thanks,
Chen

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Mar 28, 2005

Muzza

It's equivalent to proving

ab - a - b + 1 > 0
<=>
a(b - 1) - b + 1 > 0
<=>
-a(b - 1) + b - 1 < 0
<=>
(b - 1)(1 - a) < 0.

Mar 28, 2005

Chen

Argh, I was close but didn't see that! Thanks. :-)

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