[$\color{goldenrod}{\underline{\text{This}}}$ is the first part.]

Here is the “right version”. For a change we’ll take $ \mathcal{C}([a,b]) $ instead of $ \mathbb{R}^n $ (The proof structure is esentially the same). Take $ \langle f, g \rangle := \int_{a}^{b} f(t) g(t) dt $. Also for real $ p \geq 1 $, $ \lVert f \rVert_{p} := \left( \int \vert f \vert^p \right)^{\frac{1}{p}} .$

Mimicking the lemma in previous post…
$ \color{goldenrod}{\text{Lem}} $: Consider reals $ \alpha, \beta > 1 $ with $ \frac{1}{\alpha} + \frac{1}{\beta} = 1 $, and $ f,g \in \mathcal{C}([a,b]) $ with $ \lVert f \rVert_{\alpha} = \lVert g \rVert_{\beta} = 1 $.
Then $ \vert \langle f,g \rangle \vert \leq 1 $.
$ \color{goldenrod}{\text{Pf}} $: $ \vert \langle f,g \rangle \vert $ $ \leq \int \vert f \vert \vert g \vert $. As in the prev proof, $ \vert f(t) \vert \vert g(t) \vert $ $ \leq \frac{1}{\alpha} \vert f(t) \vert^{\alpha} + \frac{1}{\beta} \vert g(t) \vert^{\beta} $, so integrating gives $ \int \vert f \vert \vert g \vert \leq 1 $, as needed.

$ \color{goldenrod}{\text{Cor}} $: Suppose $ \alpha, \beta > 1 $ with $ \frac{1}{\alpha} + \frac{1}{\beta} = 1 $, and $ f,g \in \mathcal{C}([a,b]) $.
If $ \lVert f \rVert_{\alpha},\lVert g \rVert_{\beta} $ are both nonzero, notice $ u := \frac{f}{\lVert f \rVert_{\alpha}} $ and $ v := \frac{g}{\lVert g \rVert_{\beta}} $ have $ \lVert u \rVert_{\alpha} = \lVert v \rVert_{\beta} = 1 $. So $ \vert \langle u, v \rangle \vert \leq 1 $, i.e. $ \vert \langle \frac{f}{\lVert f \rVert_{\alpha}}, \frac{g}{\lVert g \rVert_{\beta}} \rangle \vert \leq 1 $, i.e. $ \vert \langle f,g \rangle \vert \leq \lVert f \rVert_{\alpha} \lVert g \rVert_{\beta} $. The inequality $ \vert \langle f,g \rangle \vert \leq \lVert f \rVert_{\alpha} \lVert g \rVert_{\beta} $ also trivially holds when of $ \lVert f \rVert_{\alpha}, \lVert g \rVert_{\beta} $ is $ 0 $. This is called Hölder’s inequality.

Continuing onto what we want…

$ \color{goldenrod}{\text{Th}} $: Let real $ p > 1 $ and $ f,g \in \mathcal{C}([a,b]) $. Then $ \lVert f + g \rVert_{p} \leq \lVert f \rVert_{p} + \lVert g \rVert_{p} $.
$ \color{goldenrod}{\text{Pf}} $: $ \lVert f + g \rVert_{p} ^p $ $ = \int \vert f+g \vert^p $ $ \leq \int \vert f+g \vert^{p-1} \vert f \vert + \int \vert f+g \vert^{p-1} \vert g \vert $ $ = \langle \vert f \vert, \vert f+g \vert^{p-1} \rangle + \langle \vert g \vert, \vert f+g \vert^{p-1} \rangle $ $ \leq \lVert f \rVert_{p} \lVert (f+g)^{p-1} \rVert_{\frac{1}{1-\frac{1}{p}}} + \lVert g \rVert_{p} \lVert (f+g)^{p-1} \rVert_{\frac{1}{1-\frac{1}{p}}}$ $ = \lVert f \rVert_{p} \lVert (f+g)^{p-1} \rVert_{\frac{p}{p-1}} + \lVert g \rVert_{p} \lVert (f+g)^{p-1} \rVert_{\frac{p}{p-1}} $ $ = \lVert f \rVert_{p} \lVert f+g \rVert_{p}^{p-1} + \lVert g \rVert_{p} \lVert f+g \rVert_{p}^{p-1}$.
So if $ \lVert f+g \rVert_{p} $ is nonzero, dividing by $ \lVert f+g \rVert^{p-1} $ gives $ \lVert f+g \rVert \leq \lVert f \rVert_{p} + \lVert g \rVert_{p} $. The inequality $ \lVert f+g \rVert \leq \lVert f \rVert_{p} + \lVert g \rVert_{p} $ also trivially holds when $ \lVert f+g \rVert_{p} $ is $ 0 $, so done.