Full expanding gives : $(x-y)(z(x^2+y^2+xy)+xy(x+y)-z^3-3xyz)+xyz(y-z) \ge 0$ which is true since $x+y\ge 2z$ and $x^2+y^2 \ge z^2 $

Ukraine 1992 huricane wrote: $3)$ should be $\frac{a^2-b^2}{c}+\frac{c^2-b^2}{a}+\frac{a^2-c^2}{b}\ge 3a-4b+c$.This inequality is simple since $\frac{a^2-b^2}{c}=\frac{(a-b)(a+b)}{c}\ge 2(a-b)$, $\frac{c^2-b^2}{a}=\frac{(c-b)(c+b)}{a}\ge 2(c-b)$ and $\frac{a^2-c^2}{b}=\frac{(a-c)(a+c)}{b}\ge a-c$.Adding those inequalities we obtain $\frac{a^2-b^2}{c}+\frac{c^2-b^2}{a}+\frac{a^2-c^2}{b}\ge 3a-4b+c$.We have equality if and only if $a=b=c$.