$\{a_n\}$ and $\{b_n\}$ are sequences of positive real numbers. Here is my current plan.

Using $c \limsup\limits_{n\rightarrow\infty} a_n = \limsup\limits_{n\rightarrow\infty} ca_n $ where c is a real number. Taking $a = \limsup\limits_{n\rightarrow\infty} a_n$ right hand side of inequality becomes $\limsup\limits_{n\rightarrow\infty} ab_n$ which is greater than LHS.

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    $\begingroup$ The plan is reasonable, but you need to give more justification for why $\limsup_{n\to\infty} ab_n$ is greater than (or equal to) the LHS. It's quite possible that $a < a_n$ for all $n$, so the inequality is not as obvious as you give it credit for. $\endgroup$ – Erick Wong Nov 8 '15 at 18:19

Hint: Using the definition of the supremum, show (or perhaps simply note) that $$ \sup_{k\geq n} a_kb_k \leq \left(\sup_{k\geq n} a_k\right) \left(\sup_{k\geq n} b_k\right) $$

  • $\begingroup$ Note that this only works if the sequences $a_n$ and $b_n$ are non-negative, which I presume is an assumption that you've neglected to mention $\endgroup$ – Omnomnomnom Nov 8 '15 at 18:22

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