These are cousins of the usual rearrangement inequality. I will prove the second one; the first one should follow using similar ideas. The proof is similar to the proof of the standard inequality in wikipedia.
Define $ F(\sigma) := \prod_{i=1}^n (a_i + b_{\sigma(i)}) . $ Let $\sigma$ be any permutation that is not the identity permutation. Then $\sigma$ contains at least one inversion; i.e., a pair $(j,k)$ such that $j < k$ and $\sigma(j) > \sigma(k)$. Let $\tau$ be the permutation $\sigma \circ (a \leftrightarrow b)$, where $(a \leftrightarrow b)$ stands for the transposition that interchanges $a$ and $b$ and keeps the remaining elements fixed.
We now claim that $F(\tau) \leqslant F(\sigma)$. To prove this, consider: $ (a_j + b_{\sigma(j)}) (a_k + b_{\sigma(k)}) - (a_j + b_{\sigma(k)}) (a_k + b_{\sigma(j)}) = (a_j - a_k)(b_{\sigma(k)} - b_{\sigma(j)}). \tag{$\ast$} $ Note that $a_j \geqslant a_k$ and $b_{\sigma(k)} \geqslant b_{\sigma(j)}$. Therefore, $(\ast)$ is nonnegative, and it follows that $F(\tau) < F(\sigma)$.
Thus, by repeatedly swapping the inverted pairs, we progressively decrease $F(\cdot)$. But since every permutation is transformed into the identity permutation by repeatedly swapping the inverted pairs, it follows that the quantity $F(\sigma)$ is minimized for $\sigma$ being the identity.