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Analysis

ian.f · 2008-12-15 14:43:34

Right lads, I'd appreciate if someone can show me how to do this one... I can see its true and think of ways of maybe saying it but I can;t string together a conclusive answer Show x^2 - bxy + y^2 > 0 for all (x,y) not= (0,0) whenever |b| < 2 Thanks

15-12-2008 2:43pm

#1

ian.f

Registered Users, Registered Users 2 Posts: 135 ✭✭

Join Date: February 2006

Posts: 110

Right lads, I'd appreciate if someone can show me how to do this one... I can see its true and think of ways of maybe saying it but I can;t string together a conclusive answer

Show x^2 - bxy + y^2 > 0 for all (x,y) not= (0,0) whenever |b| < 2

Thanks

0

Comments

#2 15-12-2008 4:01pm

Fremen

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Join Date: May 2006

Posts: 2356

Hint:

(x+y)^2 > (x-y)^2 > 0 for all x,y >0

Now mess around with adding or subtracting bxy from (x^2 + y^2). What conclusion can you draw?
A similar argument holds for other cases such as x>0, y<0

0
#3 16-12-2008 12:41pm

ian.f

Registered Users, Registered Users 2 Posts: 135 ✭✭

Join Date: February 2006

Posts: 110

Thanks fremen

0