If I know $E[\xi_{i}]=1$, what can I say about $E[\log (\xi_{i})]$?
Background: Let $(\xi_{i})_{i}$ be IID random variables with $E[\xi_{i}]=1$ where $P(\xi_{i}=1)<1$. I want to use the law of large numbers to show that $\frac{\sum\limits_{i=1}^{n}\log (\xi_{i})}{n}\to c < 0$
So I need to show that $E(\log(\xi_{i}))<0$. But all we know about the distribution of $\xi_{i}$ is that $P(\xi_{i}=1)<1$
Note: I previously proved that $X_{n}:=\prod\limits_{i=1}^{n}\xi_{i}\xrightarrow{n \to \infty} 0-$a.s, I do not know whether this helps.
Intuitively it makes sense because comparing the $\xi_{i}>1$ weight with that of $\xi_{i}<1$, moving along the $\log$ curve gives a steeper "decline" than an "increase"
You claim, that you have already proved, that $P(\Pi_{i=1}^\infty \xi_i = 0) = 1$. Thus, if you are right, then $$P(\exists n_0 \in \mathbb{N} \text{ } \forall n > n_0 \text{ } \Pi_{i=1}^n \xi_i < 1) = 1$$ That means, that $$P(\exists n_0 \in \mathbb{N} \text{ } \forall n > n_0 \text{ } \Sigma_{i=1}^n \log(\xi_i) < 0) = 1$$ This results in $$P(\exists n_0 \in \mathbb{N} \text{ } \forall n > n_0 \text{ } \frac{1}{n}\Sigma_{i=1}^n \log(\xi_i) < 0) = 1$$ And from that it follows, according to the Law of Large Numbers, that $E[\log(\xi_i)] < 0$, which seems to be the exact thing you wanted to prove.