I will use slightly different notation so that I can copy part of the text from an older answer. In that answer you can read how this is related to the existence and possible values of Banach limits.
Let $T:\ell_\infty\to\ell_\infty$ be shift-operator $T:{(x_n)}\mapsto{(x_{n+1})}$.
For any bounded sequence $x$ we define $T_n(x)=\frac{x+Tx+\dots+T^{n-1}x}n$. I.e., $T_n(x)$ is the sequence $\left(\frac{x_k+x_{k+1}+\dots+x_{k+n-1}}n\right)_{k=1}^\infty$. Let us denote $ \begin{gather*} M(x)=\lim_{n\to\infty} \limsup T_n(x) = \inf_{n\in\mathbb N} \limsup T_n(x),\\ m(x)=\lim_{n\to\infty} \liminf T_n(x) = \inf_{n\in\mathbb N} \liminf T_n(x). \end{gather*} $ The fact that the the above limits exist and that they are equal to infima can be shown using Fekete's lemma - a proof of this lemma can be found in this answer. I've added details below.
Note that $M(x)$ is the same thing as what you denoted $r(x)$ in your question.
It is easy to see, that for every $n\in\mathbb N$ and for every $x,y\in\ell_\infty$ we have $T_n(x+y)=T_n(x)+T_n(y)$. Now we get $\limsup T_n(x+y) = \limsup(T_n(x)+T_n(y)) \le \limsup T_n(x)+\limsup T_n(y)$ from the subadditivity of limit superior.
Now from the basic properties of limit you get $M(x+y) = \lim_{n\to\infty} \limsup T_n(x+y) \le \lim_{n\to\infty} (\limsup T_n(x) +\limsup T_n(y))= \lim_{n\to\infty} \limsup T_n(x) + \lim_{n\to\infty} \limsup T_n(y) = M(x)+M(y).$
(Probably it would be possible to get the required result with $\inf$ instead of $\lim$, but I think this way the solution is nicer.)
Now I get back to the fact that the both expressions (the one using $\lim$ and the one using $\inf$ are the same.)
A sequence $(a_n)$ is called subadditive if for any $m,n\in\mathbb N$ $a_{n+m}\leq a_n+a_m.$
Fekete's lemma. For every subadditive sequence $(a_n)$, the limit $\lim\limits_{n \to \infty} \frac{a_n}{n}$ exists and is equal to $\inf \frac{a_n}{n}$. (The limit may be $-\infty$.)
So to apply Fekete's lemma we need to show that $a_n=\limsup_k (x_k+x_{k+1}+\dots+x_{k+n-1})$ is a subadditive sequence. It suffices to notice that $$a_{m+n} = \limsup_{k\to\infty} (x_k+x_{k+1}+\dots+x_{k+n-1}+x_{k+n}+\dots+x_{k+n+m-1})\le \limsup_{k\to\infty} (x_k+x_{k+1}+\dots+x_{k+n-1})+ \limsup_{k\to\infty}(x_{k+n}+\dots+x_{k+n+m-1}) = a_n+a_m.$$