Assume $V$ and $W$ are complex inner product spaces and $A,B:V \rightarrow W$ are linear maps. If $\|A+B\|=\|A\|+\|B\|$ Then $\|A+3B\|=\|A\|+\|3B\|$.
This is from an old linear algebra exam and they did not mention about the norm. but it is probably the operator norm.
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In any normed linear space $\|x+y\|=\|x\|+\|y\|$ implies that $\|sx+ty\|=s\|x\|+t\|y\|$ for all $s,t \geq 0$. See An equality in normed space
This proof is independent of the norm that is given!
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In general, for any norm induced by an inner product, equality holds in the triangle inequality $\|x+y\|\le\|x\|+\|y\|$ if and only if one of $x,y$ is a nonnegative multiple of the other. More specifically, $$\|x+y\|^2=\|x\|^2+2\Re\langle x,y\rangle+\|y\|^2\le\|x\|^2+2\|x\|\|y\|+\|y\|^2=(\|x\|+\|y\|)^2.\tag{1}$$ If equality holds in $(1)$, then $\Re\langle x,y\rangle=\|x\|\|y\|$ and hence by Cauchy-Schwarz inequality, $x$ and $y$ must be parallel vectors. However, if $y=qx\ne0$ for some $q<0$ (the case $x=qy\ne0$ for some $q<0$ is similar), then $\|x+y\|=|1+q|\|x\|<(1-q)\|x\|=\|x\|+\|y\|$. Therefore, equality holds in $(1)$ only when one of $x$ and $y$ is a nonnegative multiple of the other, and obviously $\|x+y\|=\|x\|+\|y\|$ does hold in this case.
Now the statement in your question follows immediately, because one of $A,B$ is a nonnegative multiple of the other.
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It is true even if the norm is not induced from an inner product!
Just using the properties of the norm $\|\cdot\|$ and the fact that $\|A+B\|=\|A\|+\|B\|$, we obtain
$$ \|A\| +3\|B\| =3(\|A\| +\|B\|) -2\|A\| =3\|A+B\| -2\|A\| \\ \le \|3A +3B -2A\|=\|A +3B \|\le \|A\|+\|3B\| =\|A\| +3\|B\| $$ Hence $$ \|A+3B\|=\|A\| +3\|B\|. $$
In a complex inner product space, $\|u\|^2=\langle u, u\rangle$ implies
$$ \|v+w\|^2=\|v\|^2+2\mathrm{Re}\langle v,w\rangle+\|w\|^2. $$
This is a so-called polarization identity. So if we assume
$$ \|v+w\|=\|v\|+\|w\|, $$
we get $\mathrm{Re}\langle v,w\rangle=\|v\|\|w\|$. Therefore
$$ \begin{array}{ll} \|v+3w\|^2 & =\|v\|^2+2\mathrm{Re}\langle v,3w\rangle+\|3w\|^2 \\ & = \|v\|^2+6\mathrm{Re}\langle v,w\rangle+\|3w\|^2 \\ & = \|v\|^2+6\|v\|\|w\|+\|3w\|^2 \\ & = \|v\|^2+2\|v\|\|3w\|+\|3w\|^2 \\ & = (\|v\|+\|3w\|)^2. \end{array} $$
The claim follows, since complex matrices form a complex inner product space (with the so-called Frobenius aka Hilbert-Schmidt norm deriving from the inner product $\langle A,B\rangle=\mathrm{tr}(A^{\dagger}B)$).
Moreover, since $|\langle v,w\rangle|\le \|v\|\|w\|$, the only way $\mathrm{Re}\langle v,w\rangle=\|v\|\|w\|$ is possible is if there is an actual equality $\langle v,w\rangle=\|v\|\|w\|$. This is only possible if the perpendicular component of $v$ with respect to $w$ (or vice-versa) is $0$ (exercise), so $v,w$ are parallel, so wlog $v=\lambda w$ for some scalar $\lambda$, but then $\langle v,w\rangle=\|v\|\|w\|$ implies $\lambda$ is a positive real.