For $a,b,c>0$, prove that $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c+\frac{4(a-b)^2}{a+b+c}$
from your last one:
LHS$\ge (|a-b|+|b-c|+|c-a|)^2 \ge (|a-b|+|b-c+c-a|)^2=4(a-b)^2$
For $a,b,c>0$, prove that $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c+\frac{4(a-b)^2}{a+b+c}$
from your last one:
LHS$\ge (|a-b|+|b-c|+|c-a|)^2 \ge (|a-b|+|b-c+c-a|)^2=4(a-b)^2$