Proving ab - a - b + 2 > 1 for a > 1 and b > 1

[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

 

[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.

 

[Chen]

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