Do I have to prove these statements separately? If I proved:
$$\limsup\limits_{n\rightarrow \infty}(a_n+b_n)=\limsup\limits_{n\rightarrow \infty}a_n+\limsup\limits_{n\rightarrow \infty}b_n$$
Do I still need to prove: $$\liminf\limits_{n\rightarrow \infty}(a_n+b_n)=\liminf\limits_{n\rightarrow \infty}a_n+\liminf\limits_{n\rightarrow \infty}b_n$$ Or the first imply the latter?
Kind regards
 A: What is true is that $\limsup_{n\to\infty} (a_n + b_n) \le \limsup_{n\to\infty} a_n +
\limsup_{n\to\infty} b_n,$ as long as the RHS is not $\infty-\infty.$ As for the inequality in the comments,  $\sup(-A)=-\inf A$ is a general result. You just need to be a little careful with the definitions. 
First, given an arbitrary set of real numbers, $A,$ define $-A=\{-a:a\in A\}$. Now, if $A$ is bounded below, let $-\infty <\alpha=\inf A.$ Then, $\alpha\le a$ for all $a\in A,$ which implies that $-\alpha\ge -a$ for all $a\in A$, which in turn means that $-\alpha$ is an upper bound for $-A$. Because of this, $-A$ has a least upper bound $\beta<\infty.$ That is, $\beta=\sup(-A).$ Then of course, $\beta\le -\alpha.$ 
But also $\beta\ge -a$ for all $a\in A$, so $-\beta\le a$ for all $a\in A;\ $ i.e. $-\beta$ is a lower bound for $A$, and so $-\beta\le \alpha,\ $ which is the same as $\beta\ge -\alpha.$
Conclusion: $\beta=-\alpha$ and so $\sup(-A)=-\inf A.$
The case that $A$ is not bounded below is much easier. Hint: $-A$ is not bounded above. 
