I have a question about limit supremum and limit infimum. Let $\{x_n\}_{n=1}^{\infty}$ and $\{y_n\}_{n=1}^{\infty}$ be sequences of real numbers. Does the following hold:

$$ 
\limsup x_n +\liminf y_n \le \limsup\,(x_n+y_n).
$$ 

This is what I have tried but I am not quite sure if it is correct.
$\text{Fix } K&gt;1. \text{ Let }L=\inf_{1\le i \le k}y_i$. Now 
$$
\sup_{1 \le i \le k}(x_i+y_i)\ge \sup_{1\le i \le k}(x_i+L)=L+\sup_{1\le i \le k}(x_i) = 
\inf_{1\le i \le k}(y_i)+\sup_{1\le i \le k}(x_i).
$$ 
Now we take the limit as $k \rightarrow \infty$ and we have the desired result. Does this look OK?
Let $~\limsup x_n=\overline x~$ and $~\liminf y_n=\underline y~,$ then $~\overline x~,~\underline y~\in \mathbb R.$ 
Let $~\epsilon&gt;0~$ be given.
Then $~\exists~~k\in \mathbb N~$ such that $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x_n&gt;\overline x-\epsilon~/~2~,~~\forall~~n\ge k~$ and $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~y_n&gt;\underline y-\epsilon~/~2~,$ for infinitely many value of $~n~.$
Thus $~x_n+y_n&gt;\overline x+\underline y-\epsilon~,$ for infinitely many value of $~n~.$
Therefore for given $~m\in\mathbb N~,~~\exists~~n_0\ge m~$ such that $$x_{n_0}+y_{n_0}&gt;\overline x+\underline y-\epsilon$$
$$\implies \sup_{n\ge m}\left(x_n+y_n\right)\ge x_{n_0}+y_{n_0}&gt;\overline x+\underline y-\epsilon~,$$ for each $~m\in \mathbb N~.$
Hence $$\overline x+\underline y-\epsilon\le\limsup\,(x_n+y_n)$$
Since $~\epsilon &gt; 0~$ be arbitrary, $$\limsup x_n +\liminf y_n \le \limsup\,(x_n+y_n)~.$$
$$\limsup (x_n+y_n) = \lim_{n\to\infty} \left(\sup_{k&gt;n}(x_k+y_k)\right) \leq  \lim_{n\to\infty}\left(\sup_{k&gt;n}x_k+\sup_{k&gt;n}y_k\right)=\\=\limsup(x_n)+\limsup (y_n)$$

The inequality is because a property of supreme: $\sup (a+b) \leq \sup (a) + \sup (b)$.