lim sup inequality $\limsup ( a_n b_n ) \leq \limsup a_n \limsup b_n $
The basic idea is what could be called the monotonicity of $\sup$: the supremum over a set is at least as large as the supremum over a subset.
Of course, this only makes sense if the product of the $\limsup$s is not $0\cdot\infty$ or $\infty\cdot0$. We also make the assumption that $a_n,b_n\gt0$. To see that this is necessary, consider the sequences $a_n,b_n=(-1)^n-2$.
Recall the definition of $\limsup$: $$ \limsup_{n\to\infty}a_n=\lim_{k\to\infty\vphantom{d^{d^a}}}\sup_{n>k}a_n\tag{1} $$ The limit in $(1)$ exists since, by the monotonicity of $\sup$, $\sup\limits_{n>k}a_n$ is a decreasing sequence.
Furthermore, also by the monotonicity of $\sup$, if $a_n,b_n\gt0$, $$ \sup_{n>k}a_n \sup_{n>k}b_n=\sup_{m,n>k}a_nb_m\ge\sup_{n>k}a_nb_n\tag{2} $$ Taking the limit of $(2)$ as $k\to\infty$ yields $$ \limsup_{n\to\infty}a_n\limsup_{n\to\infty}b_n\ge\limsup_{n\to\infty}a_nb_n\tag{3} $$ since the limit of a product is the product of the limits.
If the limit of $a_n$ exists, we have that for any $\epsilon>0$, there is an $N$, so that $n>N$ implies $$ a_n\ge\lim_{n\to\infty}a_n-\epsilon\tag{4} $$ We are interested in small $\epsilon$, so it doesn't hurt to assume $\epsilon\lt\lim\limits_{n\to\infty}a_n$.
Thus, for $k>N$, if $a_n,b_n\gt0$, $$ \sup_{n>k}a_nb_n\ge\left(\lim_{n\to\infty}a_n-\epsilon\right)\sup_{n>k}b_n\tag{5} $$ taking the limit of $(5)$ as $k\to\infty$ yields $$ \limsup_{n\to\infty}a_nb_n\ge\left(\lim_{n\to\infty}a_n-\epsilon\right)\limsup_{n\to\infty}b_n\tag{6} $$ Since $\epsilon$ is arbitrarily small, $(6)$ becomes $$ \limsup_{n\to\infty}a_nb_n\ge\lim_{n\to\infty}a_n\limsup_{n\to\infty}b_n\tag{7} $$ Combining $(3)$ and $(7)$ yields $$ \limsup_{n\to\infty}a_nb_n=\lim_{n\to\infty}a_n\limsup_{n\to\infty}b_n\tag{8} $$ since $\displaystyle\limsup_{n\to\infty}a_n=\lim_{n\to\infty}a_n$.
$$\{{a_m}\cdot {b_m}:m\geqslant n\}\subseteq \{{a_m}\cdot {b_k}:m,k\geqslant n\}$$
since we are pairing elements from two sets together in the first set while drawing each elements at random from two sets in the second set. By taking the supremum we have:
$$\sup\{{a_m}\cdot {b_m}:m\geqslant n\}\leqslant\sup\{{a_m}\cdot {b_k}:m,k\geqslant n\}\\=\sup\{\{{a_m}:m\geqslant n\}\cdot\sup\{\{{b_m}:m\geqslant n\}$$
which is seen by using the $\textbf{lemma}$ : $\sup (A*B)=\sup A* \sup B$ , where $(A*B)=\{a*b:a\in A,b\in B\}$.
Taking limit in the above inequality gives:
$$\lim_{n\to\infty}\sup\{{a_m}\cdot{b_m}:m\geqslant n\}\leqslant \lim_{n\to\infty}\sup\{{a_m}\cdot{b_k}:m,k\geqslant n\}\\=\lim_{n\to\infty}\sup\{\{{a_m}:m\geqslant n\} \cdot\lim_{n\to\infty} \sup\{\{{b_m}:m\geqslant n\}$$ $$Q.E.D$$
Proof of $\textbf{lemma}$: First we note that for any $x,X,y,Y\in\mathbb R$, from the inequalities $$x\leq X\\y\leq Y$$ it follows that $xy\leq XY$ if either $x\ge 0$ and $Y\ge 0$ or if $y\ge 0$ and $X\ge 0$ (a sufficient condition).
Thus, if $a\ge 0,\,\forall a\in A$ and $\sup B\ge 0$ or if $b\ge 0,\forall b\in B$ and $\sup A\ge 0$, we have $$\forall c\in A*B,\exists a\in A,b\in B,s.t.c=a\cdot b\leqslant \sup A *\sup B$$ So $A*B$ is bounded by $\sup A *\sup B$.
Now, if $a\ge 0,\,\forall a\in A$ and $\sup B> 0$ or if $b\ge 0,\forall b\in B$ and $\sup A> 0$, for any small enough $\epsilon$, we have $$\forall \varepsilon \gt 0,\exists a \in A,b \in B ,s.t.a \gt \sup A-\varepsilon ,b \gt \sup B -\varepsilon ,\\a\cdot b\gt {\sup A }\cdot {\sup B}-\varepsilon\cdot \big(\sup A+\sup B)- {\varepsilon}^{2}$$
So any number less than $\sup A *\sup B $ is not an upper bound. Thus $\sup A +\sup B $ is the least upper bound.
- This inequality (for $a_n,b_n\ge0$ and excluding the indeterminate forms $0\cdot\infty$ and $\infty\cdot0$) is a part of Problem 2.4.17 in the book Wieslawa J. Kaczor, Maria T. Nowak: Problems in mathematical analysis: Volume 1; Real Numbers, Sequences and Series, Problem 2.4.15. The problem is given on p.44 and solved on p.200-201. (AFAIK this book is also available in French and Polish.) See also this answer.
I am making this CW, feel free to add other references.