Simple inequality ( or false) Given $a$, $b$ and $c > 0$ such that $abc=1$, prove that 
$$\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b} \leq 8abc$$
I don't know if its true or not but i need help in both cases.
Thanks

For $a\geq 9$ and $b=1$ and $c=1/a$ we have $$a/(b-c)=a/(1-c)>a\geq 9.$$  $$c/(a-b)>0.$$ $$ |b/(c-a)|=b/(a-c)=1/(a-1/a)\leq 1/(a-1/9)\leq 1/(9-1/9)<1/8.$$ Then the LHS is greater than $9-1/8$ which is greater than $8$ , the RHS. 
The reverse inequality does not always hold either. When $a=2,b=1, c=1/2$ the LHS is less than $4.$
