$\frac{(x-y)(x-z)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-x)(z-y)}{z^2}\geq 0$ I want to prove that
$$\frac{(x-y)(x-z)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-x)(z-y)}{z^2}\geq 0$$
for positive numbers $x,y,z$.
I don't know how to even begin. I must say I'm not 100% certain the inequality always holds.
I tried the sort of factoring involved in proving schur's inequality, but it doesn't seem to work here. I also tried to distribute the denominators to obtain terms of form (1-y/x)(1-z/x) and then maybe substituting x/y=a, y/z=b, z/x=a etc 
 A: After replacing  $x$ on $\frac{1}{x}$ and similar we need to prove that
$$\sum_{cyc}x(x-y)(x-z)\geq0,$$ which is Schur.
A: Notice, that
$$f(x,\,y,\,z)=\frac{(x-y)(x-z)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-x)(z-y)}{z^2}= \frac{z^2y^2(x-y)(x-z)+x^2z^2(y-x)(y-z)+x^2y^2(z-x)(z-y)}{x^2y^2z^2}$$
Consider two case


*

*If $x=y=z$, then it is trivial that $f(x,\,y,\,z)=0$ .



*If $x\geq y\geq z$, (without loss of generality you can commute $x,y,z$) then $z^2y^2(x-y)(x-z) \geq0 $ and $x^2y^2(z-x)(z-y)\geq 0$ and $x^2z^2(y-x)(y-z)\leq 0$. But $|x^2y^2(z-x)(z-y)|>|x^2z^2(y-x)(y-z)|$. Therefor this sum is non-negative.


A: Let $x\geq y\geq z$.
Thus, $$\sum_{cyc}\frac{(x-y)(x-z)}{x^2}\geq\frac{(x-z)(y-z)}{z^2}-\frac{(x-y)(y-z)}{y^2}=$$
$$=(y-z)\left(\frac{x-z}{z^2}-\frac{x-y}{y^2}\right)\geq0$$
because $y-z\geq0,$ $x-z\geq x-y$ and $\frac{1}{z^2}\geq\frac{1}{y^2}.$
Done!
A: There is no loss of generality in assuming $0 < x \leq y \leq z$. Rewrite the inequality as
$$\frac{(x-y)^2+(y-z)(x-y)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-y)^2+(y-x)(z-y)}{z^2} \geq 0$$
and rearrange the terms as follows:
$$\frac{(x-y)^2}{x^2}+\frac{(z-y)^2}{z^2}+(z-y)(y-x)\left(\frac{1}{x^2}+\frac{1}{z^2}-\frac{1}{y^2}\right)\geq 0.$$
Since $(z-y)(y-x)\geq 0$ by assumption, it suffices to prove 
$$\frac{1}{x^2}+\frac{1}{z^2}-\frac{1}{y^2} \geq 0,$$
which is equivalent to 
$$\left(\frac{y}{x}\right)^2+\left(\frac{y}{z}\right)^2-1\geq0.$$
But the latter is obviously true as $y/x \geq 1$.
