# Limit inferior of product of two non negative bounded sequences

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.