Maximum of $\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}$ If $a>b>c>0$ such that $7a+8b=15c+24\sqrt[3]{abc}$, find the maximum value of
$$\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}$$
I tried to use $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge 3$ so that
$$\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}=(\frac{a}{c}+\frac{b}{a}+\frac{c}{a})-(\frac{a}{b}+\frac{b}{c}+\frac{c}{b})\le (\frac{a}{c}+\frac{b}{a}+\frac{c}{a}) -3$$
and here I am stuck. 
I don't know how to find the maximum of $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$.

Notice that:
$$\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b} = \frac{(a-b)(b-c)(a-c)}{abc}$$
Now, let $x=a-b > 0$ and $y=b-c>0$. Then $a-c=x+y$ and from the condition 
$$7a+8b=15c+24\sqrt[3]{abc}\Leftrightarrow \sqrt[3]{abc}=\frac{7x+15y}{24}$$
Therefore, using AM-GM:
$$
\begin{aligned}
\frac{(a-b)(b-c)(a-c)}{abc} &= \frac{24^3 xy(x+y)}{(7x+15y)^3}\\
&= \frac{24^3}{144}\cdot \frac{(4x)\cdot (12y)\cdot 3(x+y)}{(7x+15y)^3}\\
&\leq \frac{24^3}{144\cdot 27} \cdot \frac{[4x+12y+3(x+y)]^3}{(7x+15y)^3}\\
&= \frac{32}{9}
\end{aligned}$$
Equality occurs when $a : b : c=9 : 3 : 1$.
