Denote: $\liminf a_n = a$ and $\limsup b_n = B$. Prove: $\liminf(a_n + b_n) \le a+B$.
The proof:
Let $\varepsilon > 0$.
By definition of infimum, there's a subsequence $a_{n_k}$ such that $a_{n_k} \le a +\frac{\varepsilon}{2}$
By definition of supremum, there's a subsequence $b_{n_k}$ such that $b_{n_k} \le B +\frac{\varepsilon}{2}$
Hence, (for sufficiently large $n_k$):
$$(a_{n_k} + b_{n_k}) \le a + \frac{\varepsilon}{2} + B + \frac{\varepsilon}{2} = a+ B + \varepsilon$$
Therefore, $$\liminf (a_{n_k} + b_{n_k}) \le a+ B$$
And that's implying:
$$\color{Green}{\liminf (a_n + b_n) \le a+ B}$$
I'd be glad to get an explanation for the "Green" claim.
Is it because $a_{n_k} + b_{n_k}$ is a subsequence of $a_n + b_n$?
Proof:
Let $\varepsilon>0$. Then some $k$ exists with $n>k\Rightarrow b_{n}\leq B+\varepsilon$.
Then $a_{n}+b_{n}\leq a_{n}+B+\varepsilon$ for $n>k$ and consequently $\liminf_{n\rightarrow\infty}\left(a_{n}+b_{n}\right)\leq a+B+\varepsilon$.
This is true for any $\varepsilon>0$ and this allows the conclusion that $\liminf_{n\rightarrow\infty}\left(a_{n}+b_{n}\right)\leq a+B$.