Question

hey guys , a basic question
how to prove that (a^2)+(b^2) > a*b always

Answer 1

(a-b)^2>=0 this implies a^2+b^2-2ab>=0 this implies a^2+b^2>=2ab>ab

Answer 2

@majdnation

Answer 3

can i use that to prove (a^2)+(ab)+(b^2)>=0

Answer 4

(a^2)+(b^2) > a*b always
is not true unless there are restrictions on a and b. e.g. a=b=0 gives 0>0 which is false

Answer 5

If a and b are not 0 then (a-b) is not zer and when squared is always greater than 0 therefore (a-b)^2>0 that is a^2+b^2-2ab>0 that is a^2+b^2>2ab>ab

Answer 6

so if b!=0 , only if a=b this will be wrong
thank you all