It might sound silly, but I am always curious whether Hölder's inequality $$\sum_{k=1}^n |x_k\,y_k| \le \biggl( \sum_{k=1}^n |x_k|^p \biggr)^{\!1/p\;} \biggl( \sum_{k=1}^n |y_k|^q \biggr)^{\!1/q}
\text{ for all }(x_1,\ldots,x_n),(y_1,\ldots,y_n)\in\mathbb{R}^n\text{ or }\mathbb{C}^n.$$
can be derived from the Cauchy-Schwarz inequality.
Here $\frac{1}{p}+\frac{1}{q}=1$, $p>1$.
Best Answer
Yes it can, assuming nothing more substantial than the fact that midpoint convexity implies convexity. Here are some indications of the proof in the wider context of the integration of functions.
Consider positive $p$ and $q$ such that $1/p+1/q=1$ and positive functions $f$ and $g$ sufficiently integrable with respect to a given measure for all the quantities used below to be finite. Introduce the function $F$ defined on $[0,1]$ by $$ F(t)=\int f^{pt}g^{q(1-t)}. $$ One sees that $$ F(0)=\int g^q=\|g\|_q^q,\quad F(1)=\int f^p=\|f\|_p^p,\quad F(1/p)=\int fg=\|fg\|_1. $$ Furthermore, for every $t$ and $s$ in $[0,1]$, $$ F({\textstyle{\frac12}}(t+s))=\int h_th_s,\qquad h_t=f^{pt/2}g^{q(1-t)/2},\ h_s=f^{ps/2}g^{q(1-s)/2}, $$ hence Cauchy-Schwarz inequality yields $$ F({\textstyle{\frac12}}(t+s))^2\le\int h_t^2\cdot\int h_s^2=F(t)F(s). $$ Thus, the function $(\log F)$ is midpoint convex hence convex. In particular, $1/p=(1/p)1+(1/q)0$ with $1/p+1/q=1$ hence $$ F(1/p)\le F(1)^{1/p}F(0)^{1/q}, $$ which is Hölder's inequality $\|fg\|_1\le\|f\|_p\|g\|_p$.