I have two contradicting proofs for the property:
\begin{align}\liminf (a_nb_n) \geq (\liminf (a_n)) * (\liminf (b_n))\end{align}
where ${a_n}$ and ${b_n}$ are two bounded non-negative sequences.
(Proof 1 is given in Book: Mathematical Analysis by SC Malik and Savita Arora Edition 5 Theorem 18; Proof 2 - I thought of it)
So, I was curious to know where the flaw is in the Proof 2
Proof 1: We note that if $\liminf a_n =0$ OR $\liminf b_n =0$, then the inequality follows immediately. Therefore we assume that $\liminf a_n =a>0$ and $\liminf b_n =b>0$.
Let $\epsilon>0$ be given. Then as it is known, (for limit inferiors) there exists integers $m_1$ and $m_2$ respectively, such that $$ a_n > a - (\frac{\epsilon}{2b}), \forall n\geq m_1 \\ b_n > b - (\frac{\epsilon}{2a}), \forall n\geq m_2 $$
Therefore, for all $n \geq \max(m_1, m_2)$, we have
\begin{align}
a_nb_n >(a -\frac{\epsilon}{2b})*(b-\frac{\epsilon}{2a})=ab-(\epsilon + \frac{\epsilon^2}{4ab})> ab - \epsilon
\end{align}This implies that, lim inf($a_nb_n) \geq ab - \epsilon$
But $\epsilon > 0$ was arbitrary, hence the results follows:
\begin{align}\liminf (a_nb_n) \geq (\liminf (a_n)) * (\liminf (b_n))\end{align}
Proof 2: Since $a$ and $b$ are limit inferiors of $a_n$ and $b_n$ respectively the following holds, \begin{align} a_n < (a +\frac{\epsilon}{2b}) & \text{ for infinitely many values of n}\\ b_n < (b +\frac{\epsilon}{2a}) & \text{ for infinitely many values of n} \end{align}Considering the following property of limit inferior: \begin{align*} \text{If } \alpha \in \mathbb{R }\text{ is such that \{ } {\text{n: } a_n < \alpha} \text{ is infinite \}, then lim inf } a_n \leq \alpha. \end{align*}From (1) and (2) above, we have \begin{align*} a_nb_n < ab + \epsilon + \frac{\epsilon^2}{4ab} & \text{ for infinitely many values of n} \end{align*}Now since $\epsilon$ is taken arbitrarily, considering small $\epsilon$ and also applying above property, we see that: \begin{align}\liminf (a_nb_n) \leq (\liminf (a_n)) * (\liminf (b_n))\end{align}which is a contradiction to the proof mentioned in the first post.
The second proof states that \begin{align} a_n < (a +\frac{\epsilon}{2b}) & \text{ for infinitely many values of $n$}\\ b_n < (b +\frac{\epsilon}{2a}) & \text{ for infinitely many values of $n$} \end{align} but that are not necessarily the same values of $n$ in both sequences. Therefore the conclusion \begin{align*} a_nb_n < ab + \epsilon + \frac{\epsilon^2}{4ab} & \text{ for infinitely many values of $n$} \end{align*} is wrong. As an example consider the sequences $$ a_n = 1, 2, 1, 2, 1, 2, \ldots \\ b_n = 2, 1, 2, 1, 2, 1, \ldots $$ If $\epsilon $ is sufficiently small then \begin{align} a_n < (a +\frac{\epsilon}{2b}) & \text{ for odd $n$}\\ b_n < (b +\frac{\epsilon}{2a}) & \text{ for even $n$} \end{align} so that there is no common subsequence where both estimates hold.