a,b,c,d are real no.

(+-1)(+d²)>(ac+bd-1)²

prove:

a²+b²>1 and c²+d²>1

hey a²+b²>1 bcoz in the rhs there is a square term which will always be >0 so for lhs to b +ve it has to be as c²+d² is again sum of 2 square no which is always>0

so i have proved a²+b²>1 and c²+d²>0

now i m confused as how to prove c²+d²>1.

hope u got it....

on simplifying the equation given u get

(a-c)^2>=0 so a^2+c^2>=2ac similarly for 2bd.

now (ad-bc)^2>=0

substitute above statement

.But d^2+c^2>1 can`t be proved

Yes, it can't be proved that c²+d² > 1

Question should have been like this :