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