= Solution
We use independent symmetric <Rademacher random variables> $\varepsilon_j\in\{-1,1\}$, the intended meaning of the Bernoulli variables in this context. Without this convention the assertion need not hold: for one $0/1$ Bernoulli variable of success probability $t<1/2$ and one nonzero vector, the proposed inequality would be $\sqrt t\le\sqrt2\,t$, which is false.
Let $\Omega=\{-1,1\}^n$ with the uniform <probability measure>, put $S(\varepsilon)=\sum_j\varepsilon_jx_j$, and set $F(\varepsilon)=\|S(\varepsilon)\|$. We prove the sharp estimate rather than assuming the <Kahane-Khintchine inequality>. If $\varepsilon^{(j)}$ denotes $\varepsilon$ with its $j$th sign reversed, define
$$
\mathcal Lf(\varepsilon)=\frac12\sum_{j=1}^n\bigl(f(\varepsilon^{(j)})-f(\varepsilon)\bigr).
$$
The <Walsh characters> $w_A(\varepsilon)=\prod_{j\in A}\varepsilon_j$ form an <orthonormal basis> of scalar functions on this finite cube, and $\mathcal Lw_A=-|A|w_A$. Since $F(-\varepsilon)=F(\varepsilon)$, its coefficients for odd $|A|$ vanish. Every nonconstant remaining character has $|A|\ge2$. Expanding in this <orthonormal basis> gives the <even-function spectral gap on a hypercube>
$$
2\bigl(\mathbb EF^2-(\mathbb EF)^2\bigr)\le\mathbb E\bigl[F(-\mathcal LF)\bigr].
$$
At each $\varepsilon$, the <Hahn-Banach theorem> supplies a real norming <linear functional> $\phi$ with $\|\phi\|\le1$ and $\phi(S(\varepsilon))=F(\varepsilon)$. For a complex <Banach space> use its underlying real space. Since $\|z\|\ge\phi(z)$,
$$
F(\varepsilon^{(j)})-F(\varepsilon)\ge-2\phi(\varepsilon_jx_j).
$$
Adding gives $\mathcal LF\ge-F$. Multiplying by the nonnegative $F$ yields $F(-\mathcal LF)\le F^2$. Therefore
$$
2\bigl(\mathbb EF^2-(\mathbb EF)^2\bigr)\le\mathbb EF^2,
\qquad
\boxed{\left(\mathbb E\left\|\sum_j\varepsilon_jx_j\right\|^2\right)^{1/2}\le\sqrt2\,\mathbb E\left\|\sum_j\varepsilon_jx_j\right\|.}
$$
The constant is sharp: take $n=2$ and $x_1=x_2\ne0$ in the scalar field. The random norm is $0$ or $2\|x_1\|$, each with probability one half.
A <Banach space> has <cotype 2> if a constant $C$, independent of the length of the family, satisfies
$$
\left(\sum_j\|x_j\|^2\right)^{1/2}\le C\left(\mathbb E\left\|\sum_j\varepsilon_jx_j\right\|^2\right)^{1/2}.
$$
For $f_j\in L^1(0,1)$, the <triangle inequality> for the Euclidean norm gives
$$
\left(\sum_j\|f_j\|_1^2\right)^{1/2}
=\left\|\int_0^1(|f_j(t)|)_j\,dt\right\|_2
\le\int_0^1\left(\sum_j|f_j(t)|^2\right)^{1/2}\,dt.
$$
For real or complex scalars $a_j$, independence gives $\mathbb E|\sum_j\varepsilon_ja_j|^2=\sum_j|a_j|^2$. The proved <sharp Rademacher second-moment inequality>, applied pointwise to $(f_j(t))$, and <Fubini theorem> now yield
$$
\left(\sum_j\|f_j\|_1^2\right)^{1/2}
\le\sqrt2\int_0^1\mathbb E\left|\sum_j\varepsilon_jf_j(t)\right|\,dt
=\sqrt2\,\mathbb E\left\|\sum_j\varepsilon_jf_j\right\|_1
\le\sqrt2\left(\mathbb E\left\|\sum_j\varepsilon_jf_j\right\|_1^2\right)^{1/2}.
$$
Thus \b[$L^1(0,1)$ has cotype 2, with $C\le\sqrt2$].
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