Does anybody know who to prove that for any two integers, a^2 + b^2 > 2ab?

Answer 1

a2= a*a
b2 = b*b

then by logic

aa + bb > 2ab

Answer 2

You mean a²+b²≥2ab, since obviously if a=b, this inequality is not strict. In fact, this inequality holds for all real numbers, not just integers. Consider the following:

a-b is a real number, therefore (a-b)²≥0. Expanding:
a²-2ab+b²≥0
a²+b²≥2ab

With a²+b² > 2ab if a≠b.

Answer 3

Its simple
for any two numbers a, b you ALWAYS have that (a-b)^2 is >0 (that should in fact be greater than or equal to) because squares are always non negative
but if you expand then you get
a^2-2ab+b^2>0
and hence you get what you want:
a^2 + b^2 > 2ab

Answer 4

Only for two NONEQUAL integers

if a > b
(a-b)^2 >0
a^2-2ab+b^2>0
a^2 + b^2 > 2ab
a
You get the same result if b>a, (b-a)^2 >0

If a=b
a^2 + b^2 > 2ab
a^2 + a^2 .> 2aa
2a^2 > 2a^2 (clearly false)

Answer 5

Your teachers and professors are stupid beyond belief. Rather than asking you to show that for any two real numbers, a^2 + b^2 can never be less than 2ab, they ask you to prove the above which is obviously false.

A reason most students hate math is because they have incompetent teachers/professors. The problem lies mostly with teachers in most cases as I have personally observed.

My belief is that if you can't explain given knowledge, then it is doubtful that you understand it and thus you should not be teaching it.

I suggest you go back to your teacher and ask him to prove that a^2 + b^2 cannot be less than 2ab. Let him/her break his/her stupid head over this proof.

Answer 6

a^2 + b^2 > 2ab
is equivalent to
a^2 + b^2 -2ab> 0
(a-b)^2>0
but this is NOT true, since a=b would give (a-b)^2=0,
on the other hand, if you include the 0,
i.e.
a^2 + b^2 > = 2ab
then it is true, since it is equivalent to
(a-b)^2>=0, and this is always true, since a square is always positive

Answer 7

It's not true. because if a and b both = 1 you have 2>2 which is not true.

Answer 8

If a and b=0, then you get 0>0 , which is not true.

Answer 9

for any a ,b Є R
(a - b )² ≥ 0
a² - 2ab + b² ≥ 0
a² + b² ≥ 2ab