Cauchy-Schwarz inequality and Hölder's inequality 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 $\frac{1}{p}+\frac{1}{q}=1$, $p>1$.
 A: I suggest the following consideration. We will prove the above inequality for rational $p,q\in(1,\infty)$ with $\frac1p+\frac1q= 1$, and the irrational cases follow by continuity.
If $p$ and $q$ are rational, let $p=\frac ab$ and $q=\frac ac$ with $b+c=a$ and $2^m\ge a$. Now by induction
$$ \sum |x^{(1)}\dots x^{(2^m)}| \le \left( \sum |x^{(1)}\cdots x^{(2^{m-1})}|^2 \right)^{\frac12}\cdot\left( \sum |x^{(2^{m-1}+1)}\cdots x^{(2^{m})}|^2  \right)^{\frac12}$$
$$\le \left( \sum |x^{(1)}|^{2^m} \right)^{\frac1{2^m}}\cdots \left( \sum |x^{(2^m)}|^{2^m} \right)^{\frac1{2^m}}$$
where $x^{(i)}$ are sequences of length $n$, whose indices are omitted. By plugging in
$$x^{(1)}= \dots= x^{(b)}= x^{\frac a{b2^m}},\quad x^{(b+1)}= \dots= x^{(b+c)}= y^{\frac a{c2^m}},\quad x^{(b+c+1)}= \dots = x^{(2^m)}=(xy)^{\frac1{2^m}}$$
we get 
$$\sum |xy|= \sum |x^{\frac a{2^m}}y^{\frac a{2^m}}(xy)^{\frac {2^m-a}{2^m}}| \le  \left( \sum |x|^{p} \right)^{\frac{b}{2^m}}\cdot \left( \sum |y|^{q} \right)^{\frac c{2^m}}\cdot \left( \sum |xy| \right)^{\frac {2^m-a}{2^m}}$$
implying the inequality we aimed for by dividing the third term of the RHS, and putting the inequality to the $\frac{2^m}{a}$-th power.
A: 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 $\|fg\|_1\le\|f\|_p\|g\|_p$.
