<s>[INST] Seeking nice proof of $ \frac {(ab-1)(bc-1)(ca-1)} {(a^2-1)(b^2-1)(c^2-1)} < \frac 9 8$ 
The inequality $$ 1 < \frac {(ab-1)(bc-1)(ca-1)} {(a^2-1)(b^2-1)(c^2-1)} < \frac 9 8 $$ is true if $3 \le a<b<c$.

The lower bound is quite easy.
For the upper bound let $a=x+3,b=y+3,c=z+3$ and expand 9 $\times$ denominator $-$ 8 $\times$ numerator to get a positive constant and 20 terms each a positive coefficient times some power of $x$, $y$ and $z$.
Question: Is there a "nicer" proof?
 [/INST] Per your request, for the right inequality:
Using $3 \le a < b < c$, we have
$$ab - \frac{(bc - 1)(ca - 1)}{c^2 - 1} = \frac{bc + ac - ab - 1}{c^2 - 1} > 0.$$
Using $ab > \frac{(bc - 1)(ca - 1)}{c^2 - 1} > 0$, it suffices to prove that
$$\frac{ab - 1}{(a^2 - 1)(b^2 - 1)} \cdot ab \le \frac{9}{8}$$
or
$$a^2b^2 - 9a^2 + 8ab - 9b^2 + 9 \ge 0$$
or
$$(a^2 - 9)(b^2 - 9) + 8(ab - 9) \ge 0$$
which is true.
We are done.
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