Triangle inequality fails in $L^{1,\infty}$ It can be proved that $\forall\varepsilon>0$
there exists $C(\epsilon)>0$ such that
 for all $f,g\in L^{1,\infty}(\Bbb R^n)$ we have that
$$
||f+g||_{1,\infty}\le(1+\varepsilon)||f||_{1,\infty}+C(\varepsilon)||g||_{1,\infty}$$
for example $C(\epsilon)=1+\frac1{\varepsilon}$ works.
I have some problem in proving this inequality fails for $\varepsilon=0$. I should prove that for every $C>0$ there exist $f_C,g_C\in L^{1,\infty}(\Bbb R^n)$ such that 
$$
||f+g||_{1,\infty}>||f||_{1,\infty}+C||g||_{1,\infty}
$$
 but it seems really hard. I thought I can take some sequence of functions but I can't put this idea into concrete computations.
Can someone help me?
EDIT:
Given $f:\Bbb R^n\to\Bbb R$ measurable we define, for $\alpha>0$, $\lambda_f(\alpha):=|\{x\in\Bbb R^n\;:\;|f(x)|>\alpha\}|$ (given a subset $E\subseteq\Bbb R^n$, we set |E| to denote its Lebesgue measure).
Then we set
$$
L^{1,\infty}(\Bbb R^n):=\{f:\Bbb R^n\to\Bbb R\;\mbox{measurable}\;:\exists C>0\;\;\mbox{s. t.}\;\; \lambda_f(\alpha)\le\frac C{\alpha}\;\forall \alpha>0\}\;\;\;.
$$
Finally we set $||f||_{1,\infty}$ as the infimum of such $C$'s.
 A: Obviously, $||f||_{1,\infty} = \sup_{\alpha>0} \alpha \lambda_f(\alpha)$. For $n=1$ and a monotonic $f$ this is the area of the largest rectangle under the graph of $f$. The idea is thus the following. Take $f$ such that this rectangle has a large horizontal side. We add a small $g$, which bumps $f$ at the point where the rectangle touches the graph, which increases the area considerably.
Let $n=1$, $K>0$ be small, and  $$f(x) = \frac1x \mathbf{1}_{(0,1)}(x),\quad g(x) = \frac{K}{1-x}\mathbf{1}_{(0,1)}(x).$$ Clearly, $||f||_{1,\infty}  = 1$, $||g||_{1,\infty}  = K$. It is easy to see that the minimum of $f+g$ on $(0,1)$ is attained at $x=(1+\sqrt{K})^{-1}$ and is equal to $m = (1+\sqrt{K})^2$. Therefore, 
$$
||f+g||_{1,\infty} \ge  m \lambda_{f+g}(\alpha)(m) =m\ge 1+\sqrt{K}.
$$
Taking $K<1/C^2$, we get
$$
||f+g||_{1,\infty} >  1+CK = ||f||_{1,\infty}+C||g||_{1,\infty},
$$
as required.
A: Here's an example for $\mathbb{R}$, though it can likely be generalized to higher dimensions.
Fix an integer $n\geqq 2$, and define $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=\frac{n}{i}$ if $x\in [\frac{i-1}{n},\frac{i}{n})$, $1\leq i\leq n$, and $f(x)=0$ otherwise. Then $||f||_{1,\infty}=1$.
Define a map $T:\mathbb{R}\to\mathbb{R}$ by $T(x)=x+\frac{1}{n}$ (mod 1) if $x\in[0,1)$, and $T(x)=x$ otherwise. For $1\leq j\leq n$, let $f_j(x)=f(T^jx)$.
Then $||f_j||_{1,\infty}=1$ for $1\leq j\leq n$, but $\sum_{j=1}^nf_j$ is equal to the constant function $n\sum_{j=1}^n\frac{1}{j}$ on $[0,1)$, hence
$$ \Big|\Big|\sum_{j=1}^nf_j\Big|\Big|_{1,\infty}=n\sum_{j=1}^n\frac{1}{j} $$
There is a constant $c>0$ such that $n\sum_{j=1}^n\frac{1}{j}\geq cn\log n$ for all $n\geq 2$, hence we have shown that for each $n\geq 2$ there exist $f_1,\dots,f_n$ with $||f_j||_{1,\infty}=1$ such that
$$ \Big|\Big|\sum_{j=1}^nf_j\Big|\Big|_{1,\infty}\geq cn\log n$$
If the triangle inequality holds with some constant $C$, then for any functions $g_1,\dots,g_n$ with $||g_j||_{1,\infty}=1$ we have
$$||g_1+\dots+g_n||_{1,\infty}\leq ||g_1+\dots+g_{n-1}||_{1,\infty}+C\leq\dots\leq 1+(n-1)C$$
But for any constant $C$ we can choose an $n$ sufficiently large that 
$$ cn\log n>1+(n-1)C$$
hence the triangle inequality cannot hold.
A: I found an answer by my own.
We want to prove that the statement
$$
\exists C>0\;\;:||f+g||_{1,\infty}\le||f||_{1,\infty}+C||g||_{1,\infty}\;\;\;\forall f,g\in L^{1,\infty}(\Bbb R^n)
$$
is false (we should in fact expect this: looking at PART(A), we see that $C(\varepsilon)>0$ gets bigger and bigger as $\varepsilon$ comes closer to $0$, ). Thus we must prove that
$$
\forall C>0\;\;\exists f_C,\;g_C\in  L^{1,\infty}(\Bbb R^n)\;\;:\;\; ||f_C+g_C||_{1,\infty}>||f_C||_{1,\infty}+C||g_C||_{1,\infty}\;\;.
$$
I took $n=1$ and I searched for two sequences of functions $\{f_k\}_k,\{g_k\}_k\subseteq L^{1,\infty}(\Bbb R)$ such that
\begin{equation}\label{4}
||f_k+g_k||_{1,\infty}>||f_k||_{1,\infty}+C||g_k||_{1,\infty}
\end{equation}
be true for some $k=k(C)$; in this way I would have finished.
Observe now that the above inequality is equivalent to 
$$
\frac{||f_k+g_k||_{1,\infty}}{||g_k||_{1,\infty}}>\frac{||f_k||_{1,\infty}}{||g_k||_{1,\infty}}+C
$$
(we search for a $\{g_k\}_k$ such that $||g_k||_{1,\infty}\neq0\;\;\;\forall k$; otherwise if $||g_k||_{1,\infty}=0$ for a finite number of $k$'s or for infinitely many $k$'s but not definitively, we just drop these $g_k$ from the sequence; if otherwise $g_k=0$ a.e. definitively, then  the above inequality becomes $||f_k||_{1,\infty}>||f_k||_{1,\infty}\;\;k\ge\bar k$, which is false).
Thus finding  $\{f_k\}_k,\{g_k\}_k\subseteq L^{1,\infty}(\Bbb R)$ such that
\begin{equation}\label{5} 
\left\{
\begin{array}{ll}
\limsup_k\frac{||f_k+g_k||_{1,\infty}}{||g_k||_{1,\infty}}=+\infty\\
\limsup_k\frac{||f_k||_{1,\infty}}{||g_k||_{1,\infty}}<+\infty\\
\end{array}
\right.
\end{equation}
would allow us to conclude.
\newline
\newline
I began to try with different functions on different domains, until the role of every component I was using became clear; then I could make an heuristic argument which finally gave two possible candidates for the desired sequences.
So let's $n=1$ and consider
$$
f_k(x):=-ke^x\chi_{]0,\sqrt k[}(x),\;\;\;\;\;\;\;g_k(x):=ke^x\chi_{]0,k[}\;\;,\;\;k\ge1
$$
from which
$$
f_k(x)+g_k(x)=ke^x\chi_{[\sqrt k,k[}(x)\;\;.
$$
Observe now, that, for a measurable function $f:\Bbb R^n\to\Bbb R$ we have that
$$
||f||_{1,\infty}:=\inf\left\{D>0\;:\;\lambda_f(\alpha)\le\frac D{\alpha}\;\forall\alpha>0\right\}=\sup_{\alpha>0}\{\alpha\lambda_f(\alpha)\}\;.
$$
We will use this last characterization of the $1$-weak norm in the following.
Now:
\begin{align*}
\lambda_{f_k}(\alpha)
&=
\left\{
\begin{array}{lll}
0\\
\sqrt k-\log\left(\frac{\alpha}{k}\right)\\
\sqrt k
\end{array}
\right.
\left.
\begin{array}{ccc}
\alpha\ge ke^{\sqrt k}\\
k\le\alpha<ke^{\sqrt k}\\
0<\alpha<k
\end{array}
\right.\\
&=\left(\sqrt k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[k,ke^{\sqrt k}[}(\alpha)+\sqrt k\chi_{]0,k[}(\alpha)
\end{align*}
from which we get
$$
||f_k||_{1,\infty}=\sup_{\alpha>0}\left\{\alpha\left(\sqrt k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[k,ke^{\sqrt k}[}(\alpha)+\alpha\sqrt k\chi_{]0,k[}(\alpha)\right\}=k\sqrt k
$$
in fact if we set $h(\alpha):=\alpha\left(\sqrt k-\log\left(\frac{\alpha}{k}\right)\right)$ we have $h'(\alpha)=\sqrt k-\log\left(\frac{\alpha}{k}\right)-k$ which is negative when $\alpha\in [k,ke^{\sqrt k}[$, thus here $h$ decreases, thus the $\sup$ above follows.
Let's now look at $g_k$:
\begin{align*}
\lambda_{g_k}(\alpha)
&=
\left\{
\begin{array}{lll}
0\\
k-\log\left(\frac{\alpha}{k}\right)\\
k
\end{array}
\right.
\left.
\begin{array}{ccc}
\alpha\ge ke^{k}\\
k\le\alpha<ke^{k}\\
0<\alpha<k
\end{array}
\right.\\
&=\left(k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[k,ke^{k}[}(\alpha)+k\chi_{]0,k[}(\alpha)
\end{align*}
from which we get
$$
||g_k||_{1,\infty}=\sup_{\alpha>0}\left\{\alpha\left(k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[k,ke^{k}[}(\alpha)+\alpha k\chi_{]0,k[}(\alpha)\right\}=k^2
$$
because, with the same argument as above, $h(\alpha):=\alpha\left(k-\log\left(\frac{\alpha}{k}\right)\right)$ is decreasing in $[k,ke^k[$.\
Finally let's work with the sum $f_k+g_k$:
\begin{align*}
\lambda_{f_k+g_k}(\alpha)
&=
\left\{
\begin{array}{lll}
0\\
k-\log\left(\frac{\alpha}{k}\right)\\
k-\sqrt k
\end{array}
\right.
\left.
\begin{array}{ccc}
\alpha\ge ke^{k}\\
ke^{\sqrt k}\le\alpha<ke^{k}\\
0<\alpha<ke^{\sqrt k}
\end{array}
\right.\\
&=\left(k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[ke^{\sqrt k},ke^{k}[}(\alpha)+(k-\sqrt k)\chi_{]0,ke^{\sqrt k}[}(\alpha)
\end{align*}
from which we get
$$
||f_k+g_k||_{1,\infty}=\sup_{\alpha>0}\left\{\alpha\left(k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[ke^{\sqrt k},ke^{k}[}(\alpha)+\alpha(k-\sqrt k)\chi_{]0,ke^{\sqrt k}[}(\alpha)\right\}=ke^{\sqrt k}(k-\sqrt k)
$$
again observing that $h(\alpha):=\alpha\left(k-\log\left(\frac{\alpha}{k}\right)\right)$ is decreasing in $[ke^{\sqrt k},ke^k[$.
Thus, such $f_k$'s and $g_k$'s verify the above $\limsup$'s, so we have done.
