Prove $\limsup\limits_{n \to \infty} (a_n+b_n) \le \limsup\limits_{n \to \infty} a_n + \limsup\limits_{n \to \infty} b_n$ I am stuck with the following problem.

Prove that $$\limsup_{n \to \infty} (a_n+b_n) \le \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n$$

I was thinking of using the triangle inequality saying $$|a_n + b_n| \le |a_n| + |b_n|$$ but the problem is not about absolute values of the sequence.
Intuitively it's clear that this is true because $a_n$ and $b_n$ can "reduce each others magnitude" if they have opposite signs, but I cannot express that algebraically...
Can someone help me out ?
 A: Hint:
Given two sequences, $\;\displaystyle \{a_n\}_{n \in \Bbb N},\;$ $\,\{b_n\}_{n \in \Bbb N},\;$ 
and given the definition of the supremum of a sequence, we can see that for every $k\geq n$, $$(a_k + b_k) \;\; \leq \;\;\sup_{k\geq n} a_k + \sup_{k\geq n} b_k\,.$$
Now how does this imply that $$\lim_{n\to \infty} \sup(a_n + b_n) \;\; \leq \;\; \lim _{n \to \infty} \sup a_n + \lim_{n\to \infty} \sup b_n\quad ?$$
Added: see Definition 3.16, Theorem 3.17, 3.19: perhaps you'd prefer to use the notation used there. 
A: Define for all natural numbers $k$: $A_k = \sup\{ a_n: n \ge k \}$, $B_k = \sup\{ b_n: n \ge k \}$ (where $A_k, B_k \in \mathbb{R} \cup \{+\infty\}$, (They are decreasing because for larger $k$ we take the $\sup$ of fewer terms), so that by definition $\limsup_{n \to \infty} a_n = \lim_{k \to \infty} A_k$ and similarly for $B_k$ and $\limsup_{n \to \infty} b_n$.
Also we consider the $C_k = \sup \{ (a_n + b_n) : n \ge k \}$, so that $\lim_{k \to \infty} C_k = \limsup_{n \to \infty} (a_n+b_n)$.
Now, fix an index $k$, then for all $n \ge k$ we have $a_n + b_n \le A_k + B_k$, because we estimate $a_n$ by the supremum of all terms of $(a_n)$ with $n \ge k$ and likewise for the $b_n$. As (for fixed $k$) the right hand side is fixed:
$$C_k = \sup \{ (a_n + b_n : n \ge k \} \le A_k + B_k\mbox{.}$$
This holds for all $k$, so we take the $\inf$ or $\lim$ on both sides as $k$ tends to infinity, and this preserves the inequality and we are done. 
A: This question occurs in Rudin's PMA, page 78, Exercise 5.  Below is my trial of proof. Is it right? Any comments are welcome!
5.  $\quad $For any two real sequences $\{a_n\}, \{b_n\},$  prove that 
    \begin{gather*}
  \limsup_{n\to\infty}(a_n+b_n)\leq \limsup_{n\to\infty} a_n+\limsup_{n\to\infty} b_n,
 \end{gather*}
    provided the sum on the right is not of the form $\infty-\infty.$
proof:   $\quad$  Put $a^*=\limsup\limits_{n\to\infty} a_n,  b^*=\limsup\limits_{n\to\infty} b_n$ and $c^*=\limsup\limits_{n\to\infty} (a_n+b_n).$ We shall show $c^*\leq a^*+b^*.$
Since it is excluded that the right is of the form $\infty-\infty,$ if  one of $a^*$ and $ b^*$ is $-\infty,$ then the assertion holds.  So we assume that $a^*, b^*>-\infty.$   And if one of $a^*$ or $b^*$ is $+\infty,$ then $c^*=a^*+b^*,$ so we we  need to consider the case $a^*, b^*\in\mathbb{R}.$ 
By Theorem 3.17 (see Page 56 of Rudin's PMA),  for every $\epsilon>0$ there exists $N_1, N_2$ such that 
        \begin{gather*}
   a_n<a^*+\frac{\epsilon}{2},\qquad \forall n>N_1,\\
   b_n<b^*+\frac{\epsilon}{2},\qquad \forall n>N_2.
  \end{gather*}
        Hence we have 
        \begin{gather*}
   a_n+b_n<a^*+b^*+\epsilon,\qquad \forall n>\max\{N_1, N_2\}.
  \end{gather*}
        Then, for every convergent subsequence $a_{\sigma(n)}+b_{\sigma(n)}$ of sequence $(a_n+b_n)_{n\in\mathbb{N}},$ we have
        \begin{gather*}
   \lim_{n\to\infty} (a_{\sigma(n)}+b_{\sigma(n)})\leq a^*+b^*+\epsilon.
  \end{gather*}
        Thus we see that $a^*+b^*+\epsilon$ is an upper bound of the set $C$ of subsequential limits of $(a_n+b_n)_{n\in\mathbb{N}}.$ Hence
        $c^*=\sup C\leq a^*+b^*+\epsilon.$
By the arbitrariness of $\epsilon>0,$ it follows that $c^*\leq a^*+b^*.$
$\Box$
A: Because the sup of a set is no less than the sup of a subset 
$$
\begin{align}
\sup_{k\ge n}a_k+\sup_{k\ge n}b_k
&=\sup_{\substack{j\ge n\\k\ge n}}(a_j+b_k)\\
&\ge\sup_{\substack{j\ge n\\k\ge n\\j=k}}(a_j+b_k)\\
&=\sup_{k\ge n}(a_k+b_k)\tag1
\end{align}
$$
Since $\sup\limits_{k\ge n}a_k$ is non-increasing, $\lim\limits_{n\to\infty}\sup\limits_{k\ge n}a_k$ exists. Taking $\lim\limits_{n\to\infty}$ of $(1)$ gives
$$
\limsup_{n\to\infty}a_n+\limsup_{n\to\infty}b_n\ge\limsup_{n\to\infty}(a_n+b_n)\tag2
$$
A: I happened to making solutions for PMA, so I would like to share my solution here:
If either $\limsup_{n \to \infty} a_n  = +\infty$ or $\limsup_{n \to \infty} b_n = +\infty$, there is nothing to prove. So we may assume $\limsup_{n \to \infty} a_n = A, \limsup_{n \to \infty} b_n = B$, where $A < +\infty, B < +\infty$ (but each of them can possibly take $-\infty$). 
Given $\varepsilon > 0$, there exist $N_1, N_2 \in \mathbb{N}$, such that $a_n < A + \varepsilon/2$ for all $n \geq N_1$ and $b_n < B + \varepsilon/2$ for all $n \geq N_2$. Take $N = \max(N_1, N_2)$, it follows that for all $n \geq N$, 
$$a_n + b_n < A + \varepsilon/2 + B + \varepsilon/ 2 = A + B + \varepsilon.$$
Let $n \to \infty$ in the above equation, we conclude that $\limsup_{n \to \infty} (a_n + b_n) \leq A + B + \varepsilon$. Since $\varepsilon$ is arbitrary, it follows that $\limsup_{n \to \infty}(a_n + b_n) \leq A + B$, proving the result.
