In my real analysis class, my instructor proved that $\underline{\lim} \{a_n\}+ \underline{\lim} \{b_n\} \leq \underline{\lim} (a_n + b_n)$. (Note that $\underline{\lim}$ is the limit inferior of a sequence.) but a lot of details were left out and I want to make sure that I have a correct proof and that my logic is correct.
Note that $\underline{lim} \{a_n\} = lim_{n \rightarrow \infty} \inf\{a_k|k \geq n\}$ and in my class we define $t_n =\inf\{a_k|k \geq n\}$.
My proof:
Suppose that $\underline{\lim} \{a_n\} = a$ and $\underline{\lim} \{b_n\} = b$.
First consider $\underline{\lim} \{a_n\} = a$. This means that $ \displaystyle \lim_{n \rightarrow \infty} t_n = a$. So for all $\epsilon > 0$ there exists $N_1 \in \mathbb{N}$ such that when $n \geq N_1$ we have that $ | t_n - a | < \frac{ \epsilon } { 2}$. It follows that $a - \frac{\epsilon}{2} < t_n < a_n$.
By a similar argument, there exists $N_2$ such that for all $n \geq N_2$, we have $b - \frac{\epsilon}{2} < b_n$.
So if we choose $N = \max \{N_1, N_2 \}$, for all $n \geq N$ we have
$a - \frac{\epsilon}{2} < a_n$ and
$b - \frac{\epsilon}{2} < b_n$.
Combining these inequalities gives
$a_n + b_n > a+b-\epsilon$
Then since $a+b - \epsilon$ is a lower bound for the set $\{ a_n + b_n | n \geq N \}$, we must have
$\inf\{a_n + b_n | n \geq N \} \geq a+b- \epsilon$ and
$ \displaystyle \lim_{N \rightarrow \infty} \inf\{a_n + b_n | n \geq N \} \geq a+b- \epsilon$ (because the ineq. was true for all $n \geq N$).
Then since $\epsilon$ was arbitrary,
$ \underline{ lim} (a_n + b_n) \geq a+b$