<s>[INST] For $a,b,c>0$, prove that $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c+\frac{4(a-b)^2}{a+b+c}$ 
For $a,b,c>0$, prove that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c+\frac{4(a-b)^2}{a+b+c}.$$
(Source : Balkan Mathematical Olympiad 2005)

My work:
From the given inequality, we can have,
$$\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a}\ge \frac{4(a-b)^2}{a+b+c}$$
This can be re-written as,
$$(a+b+c)\left(\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a}\right)\ge 4(a-b)^2$$
Now, what am I supposed to do? I am doubtful whether the question is correct or not. Please help.
EDIT: After checking quite a few cases I conclude that the inequality is actually working. So there might not be any mistakes at all.
 [/INST] from your last one:
LHS$\ge (|a-b|+|b-c|+|c-a|)^2 \ge (|a-b|+|b-c+c-a|)^2=4(a-b)^2$
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