[Boards: 3 / a / aco / adv / an / asp / b / biz / c / cgl / ck / cm / co / d / diy / e / fa / fit / g / gd / gif / h / hc / his / hm / hr / i / ic / int / jp / k / lgbt / lit / m / mlp / mu / n / news / o / out / p / po / pol / qa / qst / r / r9k / s / s4s / sci / soc / sp / t / tg / toy / trash / trv / tv / u / v / vg / vip /vp / vr / w / wg / wsg / wsr / x / y ] [Search | Home]
Need Help
If images are not shown try to refresh the page. If you like this website, please disable any AdBlock software!

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

File: MAT300 HW1.png (21 KB, 630x198) Image search: [iqdb] [SauceNao] [Google]
21 KB, 630x198
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