Question

Is this an appropriate way to show that 2x^2 - 4xy + 4y^2 >= 0?

Answer 1

If x = y, then
2x^2 - 4y^2 + 4y^2 >= 0
2x^2 + 4y^2 >= 4y^2.. which is obvious

and
if y=x,
2x^2 - 4x^2 + 4y^2 >=0
2x^2 + 4y^2 >= 4x^2
4y^2 >= 2x^2

Answer 2

Can I suggest another method?

Answer 3

^Sure

Answer 4

2x^2 - 4xy + 4y^2
x^2 + x^2 -2*x*2y +(2y)^2
x^2+(x-y)^2

Answer 5

Now, for any values of x and y,
x^2>=0 AND (x-y)^2>=0
Thus,
x^2+(x-y)^2>=0

Answer 6

got it?

Answer 7

AND SORRY I DONT THINK WHAT U DID IS CORRECT

Answer 8

Oh ok thanks :)

Answer 9

WELCOME