schur's inequality

If $x,y,z \in R,t\in Z$then we have the Schur's inequality:
$x^t(x-y)(x-z)+y^t(y-z)(y-x)+z^t(z-x)(z-y)\geq 0$.It is very used when determining limits of sequences but also proves many classical inequalities:
For example$(a+b-c)(b-c)(a-c)+(a+c-b)(c-b)(a-b)+(b+c-a)(b-a)(c-a)\geq 0$,where a,b,c>0
Proof:$c+b-a=x,a+c-b=y,a+b-c=z \Rightarrow a-b=\frac{y-x}{2},b-c=\frac{z-y}{2},a-c=\frac{z-c}{2}$ which becomes $x(x-y)(x-z)+...\geq 0$ which is Schur's ineq for t=1

if $x+y+z=1,x,y,z>0$then $x^2+y^2+z^2\leq 3(x^3+y^3+z^3)$
Proof:From Schu'r inequality ,in particular t=1 we have $x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y) \geq 0 \Leftrightarrow x^2(y+z)+y^2(x+z)+z^2(x+y)\leq x^3+y^3+z^3+3xyz \leq 2(x^3+y^3+z^3) \Leftrightarrow x^2(1-x)+... \leq 2(x^3+y^3+z^3)$ hence the conclusion
Study:$\frac{1}{(a-b)(b-c)(c-a)}(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})\leq 0$