It seems that your inequality is just an incarnation of Hlawka's inequality
which says that for any vectors $x, y, z$ in an inner product space $V$ we have

\begin{equation*}
\|x+y\| + \|y+z\|+\|z+x\| \le \|x\|+\|y\| + \|z\| + \|x+y+z\|.
\end{equation*}

Using $x=a+b-c$, $y=a+c-b$, and $z=b+c-a$ we obtain the inequality in the OP.

**Additional remarks:**

To add some more context, please see the paper linked here, which provides quite a nice summary of work related to Hlawka's inequality, which apparently stems back to a 1942 paper of Hornich (also cited by *Zurab* below). The paper linked to above explores the interesting generalization:
\begin{equation*}
f(x+y) + f(y+z) + f(z+x) \le f(x+y+z) + f(x)+f(y)+f(z),
\end{equation*}
where $x,y,z$ may come from an Abelian group, or a linear space, or the real line---each with its own set of conditions on the mapping $f$. The functional form of Hlawka's inequality is credited to a 1978 paper of Witsenhausen.

cannotprove it using just the triangle inequality, because it fails in $\mathbb R^3$ with the $l_\infty$ norm: just take the standard basis vectors for $a,b,c$. You’ll probably need to use that $\mathbb C$ is an inner product space. $\endgroup$12more comments