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

If $a$ and $b$ are "small" and very very close together (but $a$ slightly larger), then $c$ is "large" and $c/(a-b)$ can be made arbitrarily large.  
$a/(b-c)$ is negative but a it can be a "small" negative (as $c$ is "large" compared to $a$ and $b$.  $b/(c-a)$ is positive (but small).  So that sum can be arbitrarily large.
Example:
Fix $M$ to be a "large" number.  
Let $a = 1/M$ and $b = 1/2M$ are two small numbers close together so.....
Then $c = 1/ab = 2M^2$
Then $\frac a{b-c} = \frac 1{M(\frac 1{2M} - 2M^2)}=\frac 1{\frac 12 - 2M^3} \approx 0$
And $\frac b{c-a} = \frac{1}{2M(2M^2 - \frac 1M}=\frac{1}{4M^3- 1}  \approx 0$
And $\frac c{a-b} = \frac{2M^2}{\frac {1}{2M}} =4M^3$
Then $\frac a{b-c} + \frac b{c-a} + \frac c{a-b} \approx 4M^3 > 8 = 8abc$.
$M$ doesn't even have to be that large.  Say $M=2$ so $a= 1/2$ and $b=1/4$ and $c = 8$.
Then $\frac a{b-c} = -\frac 1{2(7\frac 34)}= -\frac {2}{31}$
$\frac b{c-a} = \frac 1{4(7 \frac 12)} = \frac 1{30}$
$\frac c{a-b} = \frac 8{1/4} = 32$ and 
$-\frac {2}{31}+ \frac 1{30} + 32 = 32 - \frac {2*30 - 31}{31*30}=32 - \frac{29}{930} = 31 \frac{901}{930} > 8*\frac 12*\frac 14*8 = 8 $
