<s>[INST] 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.
 [/INST] 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.
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