$(\int f_1d\mu)^2+\cdots+(\int f_nd\mu)^2\leq(\int \sqrt{f_1^2+\cdots+f_n^2}d\mu)^2$ Let $(X, \mathfrak{B}, \mu)$ be a measurable space, possibly not $\sigma$-finite, and $f_1, \cdots, f_n \colon X\to (-\infty, +\infty)$ be integrable functions on $X$. Does $$(\int f_1d\mu)^2+\cdots+(\int f_nd\mu)^2\leq(\int \sqrt{f_1^2+\cdots+f_n^2}d\mu)^2$$ holds? (Since $\sqrt{f_1^2+\cdots+f_n^2}\leq |f_1|+\cdots+|f_n|$, note that integrand in RHS is integrable.)
My first attempt was to apply the Fubini's theorem and Cauchy-Schwarz to the LHS: 
$\begin{align}(LHS)&=(\int f_1(x)d\mu(x))(\int f_1(y)d\mu(y))+\cdots+(\int f_n(x)d\mu(x))(\int f_n(y)d\mu(y))\\&=\int f_1(x)f_1(y)+\cdots+f_n(x)f_n(y) d(\mu\otimes\mu)(x,y)\\ &\leq\int \sqrt{f_1^2(x)+\cdots+f_n^2(x)}\sqrt{f_1^2(y)+\cdots+f_n^2(y)}d(\mu\otimes\mu)(x,y)\\&=(RHS)\end{align} $
However this approach is valid only if $X$ is $\sigma$-finite.
Note that the inequation is equivalent to the following: If $f\colon X\to \mathbb{R}^n$ is integrable, $$|\int f d\mu|\leq \int|f| d\mu$$ 
 A: Here are a couple of strategies that work in general and make no use of any type of local integrability properties of the underlying measure ($\sigma$-finiteness or not).

Consider the space $L$ of functions  $f:X\rightarrow\mathbb{R}^n$ which are integrable in each component and define $\|f\|^*=\int\|f\|_2\,d\mu$, where $\|\;\|_2$ is the Euclidean norm on $\mathbb{R}^n$. This is defines a norm on $L$ since  $\|f\|^*\leq\sum^n_{k=1}\int|f|_j\,d\mu<\infty$. Also, 
$$
\int|\|f\|_2-\|g\|_2|\,d\mu\leq\int\|f-g\|_2\,d\mu=\|f-g\|^*
$$
Consider $\mathcal{E}$ the collection of (integrable) simple functions on $(X,\mathscr{B},\mu)$
and define 
$$\mathbb{R}^n\otimes\mathcal{E}=\{\sum^m_{k=1}u_k\phi_k: u_k\in\mathbb{R}^n, \phi_k\in\mathcal{E}, m\in\mathbb{N}\}$$
This space will play the role of elementary functions in the construction of the real valued integral. It is easy to check that $\mathbb{R}^n\otimes\mathcal{E}$ is dense in $(L,\|\;\|^*)$; furthermore,  any function in $\mathbb{R}^n\otimes\mathcal{E}$ can be expressed as
$$
\Phi=\sum^{M}_{j=1}v_j\mathbb{1}_{A_j}
$$
where $v_j\in \mathbb{R}^n$, $A_j\in\mathscr{B}$, $\mu(A_j)<\infty$, and $M\in\mathbb{N}$. 
Consider now the elementary integral 
$$\int\Big(\sum^m_{k=1}u_k\phi_k\Big):=\sum^m_{j=1}u_k\int\phi_k\,d\mu$$
Since $\Phi=\sum_{u\in\mathbb{R}^n}u\mathbb{1}_{\{\Phi=u\}}$ (notice that the sum over $\mathbb{R}^n$ is actually finite),
$$
\int\Phi =\sum^m_{j=1}u_j\mu(A_j)=\sum_{u\in\mathbb{R}^n}u\int\mathbb{1}_{\{\Phi=u\}}\,d\mu\tag{1}\label{one}
$$
which means that the elementary integral extended to $\mathbb{R}^n\otimes\mathcal{E}$ does not depend on any particular representation of $\Phi$. Now
$$
\Big\|\int\Phi\Big\|_2\leq\sum_{u\in\mathbb{R}^n}\|u\|_2\int\mathbb{1}_{\{\Phi=u\}}\,d\mu=\int\Big(\sum_{u\in\mathbb{R}^n}\|u\|_2\mathbb{1}_{\{\Phi=u\}}\Big)\,d\mu=\int\|\Phi\|_2\,d\mu=\|\Phi\|^*\tag{2}\label{two}
$$
$\eqref{two}$ is the inequality you are looking for but only for functions in $\mathbb{R}^n\otimes\mathcal{E}$. For all functions in $L$ one can use some density arguments.

Comments:


*

*Notice that  $\|\;\|_2$ can be replaced by $\|\;\|_p$ ($p\geq1$).

*Your problem is an example of an integral defined on vector--valued functions.

*The arguments used, with some technical additions (Daniell integration, and measurability issues) can be used to construct Bochner's integral where  $\mathbb{R}^n$ is replaced by a Banach space.

Another, much simpler solution may be obtained by applying linear functionals to the vector $\int f=\sum^n_{j=1}e_j\int f_j\,d\mu$ where $e_1,\ldots,e_n$ is the standard basis of $\mathbb{R}^n$. As above, w $\|\,\|_p$ is $p$-norm in $\mathbb{R}^n$. We use the fact that $(\mathbb{R}^n,\|;\|_p)$ and $(\mathbb{R}^n,\|\,\|_q)$ are dual to each other when $\tfrac1p+\tfrac1q=1$.
If $\Lambda:\mathbb{R}^n\rightarrow\mathbb{}$ is linear, then $\Lambda x =x\cdot u$ for some unique $u\in\mathbb{R}$. Thus
\begin{aligned}
\Lambda \Big(\int f\Big) &= u\cdot\Big(\int f\Big)=\sum^n_{j=1}u_j\int f_j\,d\mu =\int u\cdot f\,d\mu
\end{aligned}
and so, by Hölder's inequality (in $\mathbb{R}^n$)
\begin{aligned}
\left|\Lambda \Big(\int f\Big)\right|&\leq\int|u\cdot f|\,d\mu\\
&\leq\int\|u\|_q\|f\|_p\,d\mu=\|u\|_q\int\|f\|_p\,d\mu
\end{aligned}
The result than follows by taking $\sup$ over all linear functionals $\Lambda$ with functional norm $\|\Lambda\|:=\sup_{\|x\|_p=1}|\Lambda x|\leq1$, or equivalently, by taking $\sup$ over all vectors $u\in\mathbb{R}^n$ with $\|u\|_q=1$.  Thus
$$\left\|\int f\right\|_p \leq \int\|f\|_p\,d\mu$$

A: First, assume that $(X,\mu)$ is a $\sigma$ finite space. Then there exists a probability measure $\nu$ on $X$ that is equivalent to $\mu$, that is 
$$\mu =  \rho \cdot \nu $$ where $\rho>0$ is a measurable function, $\rho>0$. We have for every $f\in L^1(X, \mu)$
$$\int_X f d\mu = \int_X f \, d\, \rho \nu = \int_X \rho f\, d \nu$$
Now, let $\phi$ be a convex function on $\mathbb{R}^n$ that is also positively homogeneous ( a sublinear function). Then we have 
$$\int_X \phi( f) d \mu= \int_X \rho \phi(f) d\nu = \int_X \phi(\rho f) d\nu \ge \phi(\int_X \rho f d\nu ) = \phi( \int_X f d\mu)$$ 
The inequality above is Jensen's inequality,  for the convex functions $\phi$ and the function $L^1$ $\rho f$ on the probability space $(X,\nu)$.
We can reduce to the case $X$ $\sigma$-finite as follows: Consider $X' = \{x\in X | f(x) \ne 0\}$. Since $f$ is $L^1$, all the subsets 
$\{x |\  |f(x)|\ge 1/n\}$ are have finite measure. Hence $X'$ is $\sigma$-finite. We can reduce all our integrals to integrals over $X'$.
Now, how to find the  probability measure $\nu$ equivalent to $\mu$. Let $X= \sqcup_n X_n$ where $\mu(X_n) <\infty$.  Now, find  $\eta>0$  such that $\int_X \eta\, d\mu = 1$, for instance $$\eta=\sum_{n\ge 1}\frac{1}{2^n} \cdot \frac{\chi(X_n)}{\mu(X_n)} $$ Put $\nu = \eta \cdot \mu$. 
$\bf{Added:}$ I think the natural solution is the second one of @Oliver Diaz, let's restate it in general terms.
Consider $\|\cdot \|$ a seminorm on $\mathbb{R}^n$ (or, more generaly, a sublinear function). We want to show the inequality
$$\| \int_X f d\mu \| \le \int_X \|f\| d \mu$$
Denote by $v \colon =  \int_X f d\mu$.  By Hahn-Banach theorem, there exists a linear functional $L\colon \mathbb{R}^n \to \mathbb{R}$ such that $L(v) = \|v\|$, and $L(w)\le \|w\|$ for all $\|w\|\in \mathbb{R}^n$. We get
$$\|\int_X f d\mu \| = L(\int_X f d\mu)=\int_X L(f) d\mu \le \int_X \|f\| d\mu$$
