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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

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    If $abc=1$, then the right-hand side is just $8$. – 2017-01-04
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    Well, abc =1 so that's the same thing as proving a/(b-c) + b/(c-a) + c/(a-b) <= 8. or a(c-a)(a-b) + b(b-c)(a-b) + c(b-c)(c-a) <= 8(a-b)(b-c)(c-a) – 2017-01-04
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    The inequality is equivalent to: if $a,b,c>0$, then $$\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b} \leq 8$$ because the inequality becomes homogeneous. – 2017-01-04

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For example, take $c$ constant ($c \neq1$ and $c>0$) and write $a=b+\epsilon$ so,

$$\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b} \leq 8 \Leftrightarrow \frac{b+\epsilon}{b-c}+\frac{b}{c-b-\epsilon}+\frac{c}{\epsilon} \leq 8$$

If we make $\epsilon \rightarrow 0$ then $a \rightarrow b \rightarrow 1/\sqrt{c}$ and the left side goes to $\infty$ and then the inequality is not true.

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Putting $a = 1/2$, $b = 4/5$, $c=5/2$ into your inequality gives $2098/255 \approx 8.23 \le 8$, which is clearly false.

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    It actualky gives 8.23 – 2017-01-04
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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 $

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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 $4.$