Can Cauchy Schwarz inequality be proven using Jensen's inequality? After reading a comment on If $\mathrm{E} |X|^2$ exists, then $\mathrm{E} X$ also exists,
I wonder if Cauchy Schwarz inequality can be proven using Jensen's inequality? 

We can prove Hölder's inequality for $p\in(1,\infty)$ (Cauchy-Schwarz is the case $p=2$ of Hölder) as follows.
Let $q(1-1/p)=1$ and assume that $f,g\ge0,\|g\|_q=1$. Let $E_g=\{x\in E:g(x)\gt0\}$. Note that on $E_g$, we also have $\|g\|_q=1$.
$$
\begin{align}
\int_Efg\,\mathrm{d}x
&=\int_{E_g}fg\,\mathrm{d}x\tag{1a}\\[3pt]
&=\int_{E_g}fg^{-q/p}g^q\,\mathrm{d}x\tag{1b}\\
&\le\left(\int_{E_g}f^pg^{-q}g^q\,\mathrm{d}x\right)^{1/p}\tag{1c}\\
&\le\left(\int_Ef^p\,\mathrm{d}x\right)^{1/p}\tag{1d}\\[9pt]
&=\|f\|_p\|g\|_q\tag{1e}
\end{align}
$$
Explanation:
$\text{(1a)}$: $g=0$ on $E\setminus E_g$
$\text{(1b)}$: $q(1-1/p)=1$
$\text{(1c)}$: apply Jensen on $E_g$ with the unit measure $g^q\mathrm{d}x$
$\text{(1d)}$: $E_g\subset E$
$\text{(1e)}$: $\|g\|_q=1$
Inequality $(1)$ scales linearly in $\|g\|_q$, so we can lift the requirement that $\|g\|_q=1$.
