For $p=1$ the result follows immediately from the triangle inequality, so we may assume $p>1$ .

We have$$ |a_k+b_k|^p=|a_k+b_k| |a_k+b_k|^{p-1} \leq (|a_k|+|b_k|)|a_k+b_k|^{p-1}$$ by the triangle inequality. Therefore we have$$ |a_k+b_k|^p \leq |a_k||a_k+b_k|^{p-1} + |b_k||a_k+b_k|^{p-1}$$ Set $q=\frac{p}{p-1}$ . Then $\frac{1}{p}+\frac{1}{q}=1$ , so by the Hölder inequality we have$$ \sum_{k=0}^n |a_k||a_k+b_k|^{p-1} \leq \left( \sum_{k=0}^n |a_k|^p \right)^{\frac{1}{p}} \left( \sum_{k=0}^n |a_k+b_k|^{(p-1)q} \right)^{\frac{1}{q}}$$ $$ \sum_{k=0}^n |b_k||a_k+b_k|^{p-1} \leq \left( \sum_{k=0}^n |b_k|^p \right)^{\frac{1}{p}} \left( \sum_{k=0}^n |a_k+b_k|^{(p-1)q} \right)^{\frac{1}{q}}$$ Adding these two inequalities, dividing by the factor common to the right sides of both, and observing that $(p-1)q=p$ by definition, we have$$ \left( \sum_{k=0}^n |a_k+b_k|^p \right)^{1-\frac{1}{q}} \leq \frac{ \sum_{k=0}^n (|a_k|+|b_k|)|a_k+b_k|^{p-1} }{ \left( \sum_{k=0}^n |a_k+b_k|^{p} \right)^{\frac{1}{q}} } \leq \left( \sum_{k=0}^n |a_k|^p \right)^{\frac{1}{p}} + \left( \sum_{k=0}^n |b_k|^p \right)^{\frac{1}{p}} $$ Finally, observe that $1-\frac{1}{q}=\frac{1}{p}$ , and the result follows as required. The proof for the integral version is analogous.