If $\liminf\limits_{n\rightarrow \infty}a_n+\liminf\limits_{n\rightarrow \infty}b_n$ exists, then $\liminf\limits_{n\rightarrow \infty}(a_n+b_n)=\liminf\limits_{n\rightarrow \infty}a_n+\liminf\limits_{n\rightarrow \infty}b_n$.
My attempt:
If either $\liminf\limits_{n\rightarrow \infty}a_n=+\infty$ or $\liminf\limits_{n\rightarrow \infty}b_n=+\infty$, then there's nothing to prove, hence we assume that $\liminf\limits_{n\rightarrow \infty}a_n=A$ and $\liminf\limits_{n\rightarrow \infty}b_n=B$, where $A>+\infty$ and $B>+\infty$. Furthermore there exists $N_1, N_2 \in \mathbb{N}$ such that $a_n<A+\varepsilon /2$ and $b_n<A+\varepsilon /2$. Let $N:=\max\{N_1,N_2\}$, then $$\forall n\geq N:a_n+b_n<A+\varepsilon /2 +B+ \varepsilon /2 =A+B+\varepsilon$$
We chose $\varepsilon$ arbitrary, thus we have $\liminf\limits_{n\rightarrow \infty}(a_n+b_n)= A+B$