$\frac{(x-y)(x-z)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-x)(z-y)}{z^2}\geq 0$

Question

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

Answer 1

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.

Answer 2

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!

Answer 3

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

1. If $x=y=z$, then it is trivial that $f(x,\,y,\,z)=0$ .
   2. 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.

Answer 4

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 $y/x \geq 1$.