s^2+t^2+2st<1
(s+t)^2<1
-1<s+t<1
s+t<1
t<1-s
hence A>B
please explain this step
-1<s+t<1
how -1<s+t?
Since (s+t)^2 <1
S+t can be 0.5, -0.5, -0.999, 0.99999
If u square any of these numbers you would still get a value less than 1
Anything but not less than -1 or greater than one
Hence the value of s+t lies between -1 and 1 excluded
Hmm could it be that when you said:
(s+t+1)(s+t-1) < 0 and concluded that (s + t + 1) < 0 or (s + t - 1) < 0 that is not true?
Since (s + t + 1) < 0 doesn’t mean (s+t+1)(s+t-1) < 0 if (s+t-1) is =< 0