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
$$a_{m+n} = \limsup (x_k+x_{k+1}+\dots+x_{k+n-1}+x_{k+n}+\dots+x_{k+n+m_1})\le
\limsup (x_k+x_{k+1}+\dots+x_{k+n-1})+\limsup(x_{k+n}+\dots+x_{k+n+m_1}) = a_n+a_m.$$