Proving Inequality: Non-Negative Variables and Limitations Explained

[Demonoid]

Question:
I need to prove this inequality:
Where x,y,x are non-negative and x+z<=2:

(x-2y+z)^2 >= 4xz -8y.

My attempt:

I thought maybe choosing x as 0 and z as 0 will and then solving for y... but that only yields y+2 >= 0, which isn't really a solution, since I can't choose numbers.

all suggestions are appreciated

thanks !

 

[verty]

Consider that if y = 0, that formula becomes (x-z)^2 >= 0. That should give you a clue.

 

[Demonoid]

Consider that if y = 0, that formula becomes (x-z)^2 >= 0. That should give you a clue.

I don't get this clue How's y ever going to be 0 ?
Even if y is 0, it will be (x+z)^2 >= 4xz, which isn't very helpful...

I need to prove this:
(x-2y+z)^2 >= 4xz -8y.

using this:
x+z<=2:I can't seem to simplify (x-2y+z)^2 >= 4xz -8y enough to get x,y,z by themselves..
I expanded first:
x^2 - 4y^2 + z^2 >= 4xz - 8y.
and then got this:

(x-z)^2 - 4y^2 + 8y^2 >= 2xz and I'm stuck...

I don't know how I can use x+z<=2 ?

 

[AlephZero]

I can't seem to simplify (x-2y+z)^2 >= 4xz -8y enough to get x,y,z by themselves..
I expanded first:
x^2 - 4y^2 + z^2 >= 4xz - 8y.

That is wrong. (a-b)^2 does not equal a^2 - b^2.

 

[verty]

Keep trying, you've made mistakes.

 

[Demonoid]

Ok, I think I'm getting closer:

Here's what I've got:

(x-z)^2 - 4(yx - yz + y^2 + 2y) >=0

now I get why y = 0, would give me (x-z)^2...

but how would y be equal to 0 ?

 

[verty]

Now you need to look at that, manipulate it, and convince yourself that it is true.

Oh, I think you have a slight error with that formula.

 

[Demonoid]

(x-z)^2 - 4yx - 4yz +4y^2 + 8y >= 0

is this right ?

 

[arildno]

Let's look at it from the start, from the following expression:
$$(x-2y+z)^{2}-4xz+8y$$
Exapanding the parenthesis, and rearranging, we get:
$$(x-2y+z)^{2}-4xz+8y=x^{2}-2xz+z^{2}+4(y^{2}+(2-(x+z))y=(x-z)^{2}+4(y^{2}+Ay), A=2-(x+z)>0$$

Now, since y is non-negative, your result follows easily..