## AGM Inequality Proof [closed]

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 :)

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MO is primarily intended for research-level questions, and so I think you question would belong better on artofproblemsolving.com/Forum/index.php or math.stackexchange.com – Yemon Choi Jan 8 2011 at 1:55
Ohh thanks, didnt know that existed. – 1337holiday Jan 8 2011 at 2:05