I have been stuck on this one for hours.
Let x,y,z be non-negative real numbers.
Also we know x + z <= 2
Prove the following:
(x - 2y + z)^2 >= 4xy - 8y
Apparently this can be proven with or without AGM which is xy <= ((x + y) / 2) ^ 2
This is what i have done so far:
((x + z) - 2y)^2 >= 4xy - 8y
(x + z)^2 -4y(x + z) + 4y^2 >= 4xy - 8y #expanded the squared term keeping (x + z) as a package
(x + z)^2 -4yx - 4yz + 4y^2 >= 4xy - 8y #now we have AGM
((x + z)/2)^2 -yx -yz + y^2 >= xy -2y
Rearrange
((x + z)/2)^2 -yx -yz + 2y >= xy - y^2
This is as far as i got and we also know that the told us this: x + z <= 2
Rearranging we get:
x + z - 2 <= 0 #then multiply by y yx + yz - 2y <= 0
I am new to this proofs and if someone can guide me as to how to attempt these and whats the method to solve these question that would be really great, thanks :)
