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?

Here are two such proofs.

One standard way of proving Hölder's inequality derives it from Young's inequality, which follows from AM/GM, which can be proven from Jensen's inequality applied to the convexity of $x\mapsto e^x$. If you take that proof of Hölder and specialize it to the case $p=q=2$, you'll have a proof of Cauchy-Schwarz which is at root from convexity.
I think this one is slightly less well-known. Consider two functions $f$ and $g$ (on $[0,1]$, say, or $\{1,\dotsc,n\}$ for the sequence case). Define $$ F = \frac{f}{\|f\|+\|g\|} \quad\text{and}\quad G = \frac{g}{\|f\|+\|g\|} $$ where $\|\cdot\|$ is the $L_2$ norm. (We do not assume that it is a norm; in fact we will prove the triangle inequality.) Note that $\|F\|+\|G\|=1$, so we can use them as coefficients in a convex combination. By Jensen's inequality and the convexity of $x\mapsto x^2$, $$ (f+g)^2 = \left(\|F\|\frac{f}{\|F\|} + \|G\|\frac{g}{\|G\|}\right)^2 \le \|F\|\left(\frac{f}{\|F\|}\right)^2 + \|G\|\left(\frac{g}{\|G\|}\right)^2 = \frac{f^2}{\|F\|} + \frac{g^2}{\|G\|} $$ Integrating yields $$ \|f+g\|^2 \le \frac{\|f\|^2}{\|F\|} + \frac{\|g\|^2}{\|G\|} = \|f\|\big(\|f\|+\|g\|\big) + \|g\|\big(\|f\|+\|g\|\big) = \big(\|f\|+\|g\|\big)^2 $$ Expanding the leftmost and rightmost expressions, we get $$ \|f\|^2 + 2\langle f,g\rangle + \|g\|^2 \le \|f\|^2 + 2\|f\|\|g\| + \|g\|^2 $$ and simplifying yields Cauchy-Schwarz. (This proof readily generalizes to give Minkowski's inequality without going through Hölder; I learned it from Garling's Inequalities: A Journey into Linear Analysis (Cambridge UP, 2007).)

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 $g^q\mathrm{d}x$
$\text{(1d)}$: $E_g\subset E$
$\text{(1e)}$: $\|g\|_q=1$