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Got a counterexample?

Yeah
prove that A^2+AB+B^2=>0
A^2+AB+B^2=>0
A^2+2AB+B^2=>AB
(A+B)^2=>AB

notice that in all cases
A>0>B
B>0>A
A=0
B=0 all cases this is true
and it's trivial if AB are the same sign that A^2+AB+B^2=>0

a1) (A-B)^2 + AB >= 0 for A*B >= 0
a2) (A+B)^2-3AB >= 0 for A*B < 0
since both a1 and a2 are the same as a, right?

>>42578
in a1 and a2 square is always >= 0
in a1 AB >= 0 since AB >= 0
in a2 -3AB > 0 since AB < 0

for the different way:
(A-B)^2>=0
or A^2+B^2 - 2AB>=0

then
for AB non negative (AB>=0)
A^2 +B^2 >= 2AB >= AB

for AB non positive (AB<=0)
A^2+B^2 >= 0 >= AB for AB non positive

not really different though, sorry