Give that: $a_n,b_n>0$ both bounded.
Prove $\lim\inf a_{n}*\lim\inf b_{n}\le \lim\inf (ab) \le \lim\inf a_{n}*\lim\sup b_{n}$
I proved the left side. The right one I need to prove for $\lim\inf b_n=0$ and $\lim\inf b_n>0$( here I can use: $\lim\sup(1/a_n)=1/\lim\inf a$
Fix $\varepsilon>0$, by definition of liminf there is $N\in\mathbb{N}$ such that $$ b_n \geq \tilde{b}-\varepsilon, \quad \forall\ n\geq N,$$ where $\tilde{b}\equiv \liminf b_n$. Analogously for the $\limsup$. Note that, because the sequences $a_n,b_n$ are bounded, then $\liminf$ and $\limsup$ exist finite.
Therefore, because $a_n>0$, $$ \liminf a_n b_n \geq \liminf a_n (\tilde{b}-\varepsilon) = \liminf a_n \liminf b_n - \varepsilon \liminf a_n; $$ letting $\varepsilon\to 0$ we get the first inequality.
The second inequality is immediate. After naming $\bar{b}=\limsup b_n$, we get $$ \liminf a_n b_n \leq \liminf a_n (\bar{b}+\varepsilon) = \liminf a_n \limsup b_n + \varepsilon \liminf a_n. $$ Letting $\varepsilon\to 0$ we recover the right inequality.