= Solution
First use independent fair signs $\varepsilon_i\in\{-1,1\}$, the <Rademacher distribution> convention for <Bernoulli random variables> in this setting. Let $\Omega=\{-1,1\}^d$ with uniform <probability>, write $s(\omega)=\sum_i\omega_i a_i$ and $F(\omega)=\|s(\omega)\|$. We prove the <sharp Rademacher second-moment inequality> by a complete finite-cube calculation.
For a function on this cube define the <coordinate-flip generator on a hypercube>
$$
Lu(\omega)=\frac12\sum_{i=1}^d
\bigl(u(\omega^{(i)})-u(\omega)\bigr),
$$
where $\omega^{(i)}$ flips just coordinate $i$. The <Walsh characters> $\chi_A(\omega)=\prod_{i\in A}\omega_i$ form an <orthonormal basis>: distinct characters have expectation-zero product, and there are $2^d$ of them. Flipping a coordinate in $A$ changes its sign and other flips leave it unchanged, so $L\chi_A=-|A|\chi_A$. Expand the real function $F$ as $\sum_A\widehat F(A)\chi_A$. Since $F(-\omega)=F(\omega)$, changing variables $\omega\mapsto-\omega$ shows that $\widehat F(A)=0$ whenever $|A|$ is odd. Thus
$$
-\mathbb E[FLF]=\sum_A|A|\widehat F(A)^2
\geq2\sum_{A\ne\varnothing}\widehat F(A)^2
=2\operatorname{Var}(F).
$$
This derives the <even-function spectral gap on a hypercube>, rather than assuming it.
At each vertex with $s(\omega)\ne0$, choose a <norming functional> $\ell_\omega$ of <norm> one satisfying $\ell_\omega(s(\omega))=F(\omega)$, supplied by the <Hahn-Banach theorem>. For complex $E$ take <real parts> in the following inequalities. At a vertex with $s(\omega)=0$, use the zero functional. The supporting inequality for the <norm> gives
$$
F(\omega^{(i)})-F(\omega)
\geq\operatorname{Re}\ell_\omega
\bigl(s(\omega^{(i)})-s(\omega)\bigr).
$$
But $Ls=-s$, since every coefficient is a degree-one <Walsh character>. Summing the preceding inequalities with factor $1/2$ therefore proves $LF(\omega)\geq-F(\omega)$. As $F\geq0$,
$$
2\bigl(\mathbb EF^2-(\mathbb EF)^2\bigr)
\leq-\mathbb E[FLF]\leq\mathbb EF^2.
$$
Rearranging and taking <square roots> yields
$$
\boxed{\left\|\sum_i\varepsilon_i a_i\right\|_{L^2(E)}
\leq\sqrt2\left\|\sum_i\varepsilon_i a_i\right\|_{L^1(E)}.}
$$
No completeness of $E$ is needed. The constant is sharp: with two identical nonzero scalar coefficients, the absolute sum is zero with <probability> $1/2$ and twice the coefficient magnitude with <probability> $1/2$.
If Bernoulli instead means fair $\{0,1\}$ variables $b_i$, write $b_i=(1+\varepsilon_i)/2$ and $a=\sum_i a_i$. Add an independent fair sign $\varepsilon_0$ and apply the proved estimate to $Y=\varepsilon_0a+\sum_i\varepsilon_i a_i$. Conditional on either sign of $\varepsilon_0$, $\|Y\|$ has the same distribution as $\|a+\sum_i\varepsilon_i a_i\|$, by symmetry of the sign sum. Hence $\|Y\|_{L^p(E)}=2\|\sum_i b_i a_i\|_{L^p(E)}$ for $p=1,2$, proving the same bound in that convention.
The constant is documented in https://www.impan.pl/en/publishing-house/journals-and-series/studia-mathematica/all/109/1/108398/on-the-best-constant-in-the-khinchin-kahane-inequality[the primary sharp-constant study].
Fairness is essential. For a general $\{0,1\}$ <Bernoulli random variable> with success <probability> $0<t<1/2$, a single nonzero scalar coefficient gives $L^2/L^1=1/\sqrt t>\sqrt2$. Thus the printed term cannot be read as permitting arbitrary Bernoulli <probabilities>.
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