Given $(x_i)_{i=1}^n, (y_i)_{i=1}^n \in \mathbb{R}^n$, the Cauchy-Schwarz Inequality asserts $$\left( \sum_{i=1}^n x_i y_i \right)^2 \leq \left( \sum_{i=1}^n x_i \right)^2 \left( \sum_{i=1}^n y_i \right)^2.$$
Conspicuously absent from the Wikipedia page is a claim that $$\left( \sum_{i=1}^n x_i y_i \right)^k \leq \left( \sum_{i=1}^n x_i^k \right) \left( \sum_{i=1}^n y_i^k \right)$$ holds for $k>2$, which makes me think it's untrue (and similarly for the same formula with absolute value signs around the $x_i$'s and $y_i$'s). Indeed, we can find random counter-examples on a computer.
Question: Why doesn't Cauchy-Schwarz in $\mathbb{R}^n$ generalize to exponents $k>2$?
Can we gain any insight into why it works for $k=2$ but not for $k>2$?
Cauchy-Schwarz inequality gives an optimal comparison between the Euclidian inner product and the Euclidian norm. The generalization would compare the latter with the $\ell_k$-norm for $k\gt 2$. Since $\mathbb R^n$ is finite dimensional, we know that there is $C_n$ such that for all $x,y\in\mathbb R^n$, $$|\langle x,y\rangle|\leqslant C_n\lVert x\rVert_{\ell_k}\cdot \lVert y\rVert_{\ell_k},$$ but taking $x=y=(1,\dots,1)^t$, we can see that $n^{k-2}\leqslant C_n$, so we cannot choose $C_n=1$.