Prove $\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\ge 4+(x-y)^2$ for $4 \le x + y + z \le 5$ 
Let $x,y,z>0$, and such $$4\le x+y+z\le 5.$$
Show that
$$\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge 4+(x-y)^2.$$

It seems that the condition $\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge 4+(x-y)^2$ is maybe old, and this condition is strange?
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=586357 .
I want to use
$$\dfrac{x^2}{y}=2x-y+\dfrac{(x-y)^2}{y}.$$
If we let $$x\to x'r,y\to y'r,z\to z'r,$$then $$x'+y'+z'=4, r\in [1,\dfrac{5}{4}]$$
But I can't prove that either, because I felt I can't use the condition.
Thank you.
 A: Notes
Pardon my long, long answer (I struggle to be concise). I'll give my best to make my answer simple, clear, and easy to understand.
Because this is a rather "confusing" problem, I cannot provide an extremely concrete proof - part of this might need your intuition's help to make sense (sorry!).
EDIT: I also do realize that this answer is, in great parts, incomplete in that it does not make use of $x + y + z \ge 4$. That is a problem because we know well of many examples that don't satisfy this inequality for $x + y + z < 4$, e.g. $(1, 1, 1)$. It would really help me as well if you can find the little gap in my logic for me, thanks!
Please do correct me if you find anything in my solution that is causing a problem. This is, after all, just my weak attempt at tackling this 1k-view question.
My Work
To start with, we know we want to prove
$$\frac{x^2}y + \frac{y^2}z + \frac{z^2}x \ge 4 + (x-y)^2$$
We can multiply the entire inequality by $xyz$, and we know we can do that without needing to change the signs because we are given that $x, y, z > 0$. Thus, we get
$$x^3z + y^3x + z^3y \ge 4xyz + xyz(x^2 - 2xy + y^2)$$
$$x^3z + y^3x + z^3y \ge 4xyz + x^3yz - 2x^2y^2z + y^3xz$$
We'll be working from that. From the second given, we know that
$$x + y + z <= 5$$ 
As we also know that $x, y, z > 0$ (Note that $x, y, z \neq 0$). From some basic reasoning (or, if you insist, simple algebra), you can realize that 
$$x, y, z < 5$$
We'll then start by breaking the inequality down into parts, labeled as follows:
$$x^3z + y^3x + z^3y \ge 4xyz + x^3yz - 2x^2y^2z + y^3xz$$
$$\underbrace{x^3z + y^3x + z^3y}_A \ge \underbrace{4xyz + x^3yz + y^3xz}_B + \underbrace{(-2x^2y^2z)}_C$$
$$A \ge B + C$$
Group $A$
We can first find an inequality to describe $A$. Because we know $x, y, z < 5$, we can be completely sure that $x^3z, y^3x, z^3y < 5^4$, or $x^3z + y^3x + z^3y < 3 \cdot 5^4$. So we know:
$$A < 3 \cdot 5^4$$
$$A < 1875$$
Group $B$
To describe $B$, we do, essentially, the same thing. We break the group down into parts:
$$4xyz < 4 \cdot 5^3$$
$$x^3yz, y^3xz < 5^5$$
We sum the three to get:
$$B < 4 \cdot 5^3 + 2\cdot 5^5$$
$$B < 6750$$
Group $C$
While finding an inequality for $C$, we must remember that $C$ is negative. Keep in mind, therefore, that the signs will be flipped. We know $x^2y^2z < 5^5$. We multiply both sides by $-2$ and flip the inequality sign to get
$$-2x^2y^2z > -2\cdot 5^5$$
$$C > -6250$$
Simplifying the Problem
Alright now. With our inequalities, I think it makes sense to say that we can prove the main inequality by proving our "reworded" inequality
$$A \ge B + C$$
given the three conditions
$$A < 1875$$
$$B < 6750$$
$$C > -6250$$
Seriously, correct me if I'm wrong here. I really could be - it almost seems so confusing.
The Final Proof
Let's first plug in the inequality for $A$ and simplify.
$$1875 > A \ge B + C$$
$$1875 > B + C$$
Then, we can subtract $B$ from both sides and plug in $C$'s inequality and simplify again:
$$1875 - B > C$$
$$1875 - B > C > -6250$$
$$1875 - B > -6250$$
Lastly, we add $B$ and $6250$ to both sides to simplify:
$$1875 + 6250 > B$$
$$8125 > B$$
I would think, then, that all we need do is to show $$8125 > B$$ for all $B < 6750$. This is easily done because we know that anything less than $6750$ is also less than $8125$. So I'd think it's proved.
But again, this, for one, seems too ambiguous logically - and, all the while, too simple.
I also, again, notice that I did not make use of $x + y + z \ge 4$ - so I'm almost sure some part of this proof is drastically misled. However, for now, this is the best I can give - please, please, please do point out any mistake you find in my work - it'd help me too!
A: I cannot give a full proof but here is a piece that might take you closer to what you're looking for.
Using the equality you have given, we can transform the left-hand side of the equation to the form
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x} = 2x - y + \frac{(x-y)^2}{y} + 2y - z + \frac{(y-z)^2}{z} + 2z - x + \frac{(z-x)^2}{x} = x + y + z + \frac{(x-y)^2}{y} + \frac{(y-z)^2}{z} + \frac{(z-x)^2}{x} $$
Since we know that $x+y+z \geq 4$, we get
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x} \geq 4 +  \frac{(x-y)^2}{y} + \frac{(y-z)^2}{z} + \frac{(z-x)^2}{x}$$
So, in order to prove the given inequality, it is sufficient to prove that 
$$\frac{(x-y)^2}{y} + \frac{(y-z)^2}{z} + \frac{(z-x)^2}{x} \geq (x-y)^2$$
This inequality can be rewritten as
$$(x-y)^2 \left(\frac{1}{y} - 1 \right) + \frac{(y-z)^2}{z} + \frac{(z-x)^2}{x} \geq 0$$
Since any square $x^2 \geq 0$ and $x,y,z > 0$, this inequality certainly holds when 
$$\frac{1}{y} - 1 \geq 0$$ or in other words when $0 < y \leq 1$.
However, this is a rather strict condition that was not actually given. Additionally, the condition $x+y+z \leq 5$ was not used anywhere in here - and as @EwanDelanoy pointed out, it seems to be crucial.
