I am trying to prove this inequality for Darboux integrals of bounded functions $f,g$:
$$\underline{\int_a^b} f(x) dx + \overline{\int_a^b} g(x) dx \leqslant \overline{\int_a^b} (f(x)+g(x))dx\leqslant \overline{\int_a^b} f(x)dx + \overline{\int_a^b} g(x)dx $$
To prove the right side inequality, there are partitions $\mathcal{P}_1$ and $\mathcal{P}_2$ so that $U(f,\mathcal{P}_1) < \overline{\int_a^b} f(x)dx + \epsilon$ and $U(g,\mathcal{P}_2) < \overline{\int_a^b} g(x)dx + \epsilon$. If $\mathcal{P}$ is a refinement of $\mathcal{P}_1$ and $\mathcal{P}_2$ then $U(f,\mathcal{P}) + U(g,\mathcal{P}) \leq \overline{\int_a^b} f(x)dx + \overline{\int_a^b} g(x)dx+ 2\epsilon$. Then it can be shown that $\overline{\int_a^b} (f(x)+g(x))dx \leq U(f+g,\mathcal{P}) \leq U(f,\mathcal{P}) + U(g,\mathcal{P})$ to finish.
But I do not see how to get the left inequality.
Let $P$ be any partition of $[a,b]$. For any $\epsilon > 0$ there exists a partition $P'$ such that the lower Darboux sum of $f$ satisfies $\underline{\int_a^b} f - \epsilon < L(P',f).$ Taking the common refinement $Q = P\cup P'$, we have
$$\underline{\int_a^b} f - \epsilon < L(P',f) \leqslant L(Q,f)$$
Thus,
$$\tag{1}\underline{\int_a^b} f + \overline{\int_a^b} g- \epsilon < L(Q,f) + U(Q,g)$$
Given any subinterval $I$ of partition $Q$, we have $ \inf_{x \in I}f(x) + g(x)\leqslant f(x) + g(x) \leqslant \sup_{x \in I} [f(x) + g(x)]$, which implies $ \inf_{x \in I}f(x) + \sup_{x \in I} g(x)\leqslant \sup_{x \in I} [f(x) + g(x)]$ and
$$\tag{2}L(Q,f) + U(Q,g) \leqslant U(Q,f+g)$$
Using (1) and (2), we get
$$\underline{\int_a^b} f + \overline{\int_a^b} g- \epsilon < L(Q,f) + U(Q,g) \leqslant U(Q,f+g) \leqslant U(P,f+g),$$
where the last inequality holds because $Q$ is a refinement of $P$. Recall that the partition $P$ is arbitrary and, thus, taking the infimum over all $P$ we get $$\underline{\int_a^b} f + \overline{\int_a^b} g- \epsilon \leqslant \overline{\int_a^b} (f+g)$$
Since $\epsilon$ may be arbitrarily close to $0$ the desired inequality follows.