Theorem 12.1.6 (Minkowski’s Inequality, [6.5, Fol99]). Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, $p \in [1, \infty]$, and $f, g \in \mathcal{L}^{p}(X; E)$, then
\[\norm{f + g}_{L^p(X; E)}\le \norm{f}_{L^p(X; E)}+ \norm{g}_{L^p(X; E)}\]
Proof. If $p = 1$, then the theorem holds directly.
If $p = \infty$, then for any $\alpha > \norm{f}_{L^\infty(X; E)}$ and $\beta > \norm{g}_{L^\infty(X; E)}$, $\alpha + \beta \ge f + g$ $\mu$-almost everywhere, so
\[\norm{f + g}_{L^\infty(X; E)}\le \norm{f}_{L^\infty(X; E)}+ \norm{g}_{L^\infty(X; E)}\]
Now suppose that $p \in (1, \infty)$, then $p = q(p - 1)$, and
\begin{align*}\norm{f + g}_{E}^{p}&\le (\norm{f}_{E} + \norm{g}_{E})\norm{f + g}_{E}^{p - 1}\\ \int\norm{f + g}_{E}^{p}d\mu&\le (\norm{f}_{L^p(X; E)}+ \norm{g}_{L^p(X; E)})\braks{\int \norm{f + g}_E^{q(p - 1)}d\mu}^{1/q}\\ \int\norm{f + g}_{E}^{p}d\mu&\le (\norm{f}_{L^p(X; E)}+ \norm{g}_{L^p(X; E)})\braks{\int \norm{f + g}_E^{p}d\mu}^{1/q}\\ \braks{\int\norm{f + g}_E^pd\mu}^{1/p}&\le \norm{f}_{L^p(X; E)}+ \norm{g}_{L^p(X; E)}\end{align*}
$\square$