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You are currently reading a thread in /wsr/ - Worksafe Requests

You are currently reading a thread in /wsr/ - Worksafe Requests

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Anonymous

Need Help 2016-01-26 10:22:03 Post No. 42534

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Need Help 2016-01-26 10:22:03 Post No. 42534

[Report] Image search: [iqdb] [SauceNao] [Google]

So, am I supposed to prove that 1a is incorrect, since it's not true?

>>

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

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