Let $a>0$, $b>1$. Let $(X_k)$ be a sequence of iid random variables, with density $f(t) =\frac{a}{b} \left(\frac{b}{t}\right)^{a+1}$ for $t>b$ and 0 otherwise.
I can work out that the CDF is $F_{X_k} (x) =1-(b/x)^a$.
Let $Z_n:=\prod_{k=1}^nX_k$, and let $Z$ be any random variable. I'd like to show the inequality:
$\mathbb{P} (X_{n+1}>3/2) < \mathbb{P} (|Z_{n+1}-Z|>1/4)+\mathbb{P}(|Z_n-Z|>1/4)$.
The only thing I can think of is work out the lhs to be $(2b/3)^a$, and use sub additivity for the rhs, and hope that $\mathbb{P} ([|Z_{n+1}-Z|>1/4] \cup [|Z_n-Z|>1/4)] $ will already be greater than the lhs, but to no avail.
Presumably I should use both indepedence and the definition of $Z_{n+1}=Z_n X_{n+1}$, and perhaps $b>1$, but cannot find how.
First note that $x_k >b>1$ for all $k$ so $Z_n >1$ for all $n$.
Let $X_{n+1} >\frac 3 2 $. Then $ \frac {Z_{n+1}} {Z_n} >\frac 3 2 $ which gives $ \frac {Z_{n+1}-Z_n} {Z_n} >\frac 3 2-1=\frac 1 2 $. Hence, $Z_{n+1}-Z_n >\frac 1 2Z_n >\frac 1 2$. Can you see now why we must have either $|Z_{n+1}-Z| >\frac 1 4$ or $|Z_n-Z| >\frac 1 4$?