Is $\iiint_{[0, 1]^3} \lvert f(x)+f(y)+f(z)\rvert\, dx\, dy\, dz \ge \int_0^1 \lvert f(x)\rvert\, dx$? $\newcommand\abs[1]{\lvert#1\rvert}\newcommand\Abs[1]{\left\lvert#1\right\rvert}$Question: Let $f(x)$ be a continuous real-valued function deﬁned on the interval $[0, 1]$. Does the following inequality hold? $$\int_{0}^{1}\int_{0}^{1}\dotsi\int_0^1\int_0^1\abs{f(x_{1})+f(x_{2})+\dotsb+f(x_{n})}dx_1 \; dx_{2}\dotsm\;dx_{n} \ge \int_0^1 \abs{f(x)}dx.$$
Even the case $n=3$ would be interesting. This is inspired by the case $n=2$, which was a Putnam problem from 2003, as follows.
Theorem. Let $f(x)$ be a continuous real-valued function deﬁned on the interval $[0, 1]$. Then $$\int_0^1\int_0^1\abs{f(x)+f(y)}dx \; dy \ge \int_0^1 \abs{f(x)}dx.$$
Proof by Kent Merryfield:
Let $P$ be the subset of $[0, 1]$ on which $f\ge 0$ and $N$ the set on which $f < 0$. As is conventional, define $f^+(x) = \max(f(x), 0)$ and $f^-(x) = \max(-f(x), 0)$. Thus, $f = f^+ - f^-$, $|f| = f^+ + f^-$, and $f^+$ equals $0$ everywhere on $N$ while $f^-$ equals zero everywhere on $P$.
Then
\begin{align*}\int_0^1 \int_0^1   \abs{f(x) + f(y)}\,dx\,dy &= \int_P  \int_P   \abs{f(x) + f(y)}\,dx,dy +  \int_P  \int_N   \abs{f(x) + f(y)} \,dx\, dy  \\
                &+ \int_N  \int_P   \abs{f(x) + f(y)}\, dx \,dy+ \int_N \int_N   \abs{f(x) + f(y)}\, dx\, dy.\end{align*}
We tackle these terms one at a time.
\begin{align*}\int_P  \int_P   \abs{f(x) + f(y)}\, dx\, dy &=  \int_P  \int_P   (f(x) + f(y))\,dx \,dy\\
        &= \abs P \int_P   f(x)\,dx  + \abs P  \int_P   f(y) \,dy = 2\abs P\int_P   f^+(x) \,dx\end{align*}
where we use the notation $\abs P$ to mean the measure (total net length) of the set $P$.
Similarly, $\int_N \int_N   \abs{f(x) + f(y)} \,dx \,dy = 2\abs N\int_N   f^-(x) \,dx$.
The other two terms are equal to each other (as shown by interchanging $x$ and $y$).
\begin{align*}\int_P \int_N   \abs{f(x) + f(y)} \,dx \,dy &=  \int_P  \int_N   \abs{f^+(x) - f^-(y)} \,dx \,dy\\
&\ge \Abs{\int_P  \int_N   f^+(x) - f^-(y) \,dx \,dy}\\
&=  \Abs{ \abs N\int_P   f^+(x) \,dx -  \abs P\int_N   f^-(y) \,dy}\end{align*}
If we let $A =  \int_P   f^+(x) \,dx$,  $B =  \int_N   f^-(x) \,dx$, and $I = \int_0^1 \int_0^1   \abs{f(x) + f(y)} \,dx \,dy$, then we have found that $I \ge 2\abs P A + 2\abs N B + 2\abs{(\abs N A - \abs P B)}$.
For convenience, we now square this:
\begin{align*}I^2   &\ge 4\left[(\abs P A + \abs N B)^2 + (\abs N A - \abs P B)^2 + (\text{other positive terms})\right]\\
    &\ge 4(\abs P^2A^2 + \abs N^2B^2 + \abs N^2A^2 + \abs P^2B^2)\\
    &= 4(\abs P^2 + \abs N^2)(A^2 + B^2).\end{align*}
But for real $a$ and  $b$, $(a + b)^2 \le 2(a^2 + b^2)$ since $2(a^2 + b^2) - (a + b)^2 = (a - b)^2$.
Hence, $2(\abs P^2 + \abs N^2) \ge (\abs P + \abs N)^2 = 1^2$, since $\abs P + \abs N$ is the measure of the interval $[0, 1]$.
Also, $2(A^2 + B^2) \ge (A + B)^2 =  \left(\int_0^1  \abs{f(x)} \,dx\right)^2$. $\square$
 A: Here is the proof using the alternative route.
Let $X$, $Y$ be two independent real-valued random variables such that $EX,EY\ge 0$ and $\min(E|X|,E|Y|)=I$. We want to prove that $E|X+Y|\ge I$. Again, as in both the OP and Iosif's post, we can consider only the case when $X$ is $A$ with probability $P$ and $-B$ with probability $Q$, while $Y$ is $a$ with probability $p$ and $-b$ with probability $q$, where $A,B,a,b\ge 0$.
Then we need to show that the inequality
$$
(A+a)Pp+|a-B|pQ+|b-A|Pq+(B+b)Qq<I
$$
is impossible. So suppose it holds.
If we estimate each absolute value by what is inside it, we get
$$
(A+a)Pp+(a-B)pQ+(b-A)Pq+(B+b)Qq
\\
=(ap+bq)+(AP-BQ)(p-q)
\\=E|Y|+(EX)(p-q)<I\,,
$$
which is possible only if $p<q$.
By symmetry, we conclude that we must also have $P<Q$.
Now estimate $|a-B|\ge a-B$ and $|b-A|\ge A-b$. We obtain
$$
(A+a)Pp+(a-B)pQ+(A-b)Pq+(B+b)Qq
\\
=(AP+ap)+BQ(q-p)+bq(Q-P)<I\,.
$$
However, the last two terms are nonnegative and the condition $0\le EX=AP-BQ$ implies that $AP\ge \frac 12[AP+BQ]=\frac 12E|X|\ge\frac 12I$ and similarly for $ap$, so we run into a contradiction.
Edit Here is the direct binomial coefficient approach. I'll start where Iosif stopped though I'll denote by $A$ what he denotes by $A/p$, etc.
We need the inequality
$$
\sum_{k=0}^n{n\choose k}p^kq^{n-k}|Ak-B(n-k)|\ge Ap+Bq
$$
Fix $A+B=1$ and consider the difference as a function of $A\in[0,1]$. It is piecewise linear, so the minimum is attained at a zero of some absolute value. Clearly, the endpoints (when the random variable preserves sign) are not competitors, so it is enough to consider the case when $A=\frac{n-\ell}n, B=\frac{\ell}n$, $0<\ell<n$. Then all the absolute values except the vanishing one for $k=\ell$ are at least $A+B=1$, so the LHS is at least
$
1-{n\choose \ell}p^\ell q^{n-\ell}
$
and we want to show that
$$
1-{n\choose \ell}p^\ell q^{n-\ell}\ge Ap+Bq\,,
$$
i.e.,
$$
Aq+Bp\ge {n\choose \ell} p^{\ell}q^{n-\ell}
$$
Since $Aq+Bp\ge q^Ap^B=p^{\frac{\ell}{n}}q^{\frac{n-\ell}{n}}$, it suffices to check that
$$
{n\choose \ell} [p^{\ell}q^{n-\ell}]^{1-\frac 1n}\le 1\,.
$$
The product in brackets is maximized when $p=\frac \ell n$, $q=\frac{n-\ell}{n}$, so we want
$$
{n\choose \ell}[(\tfrac{\ell}n)^{\ell}(\tfrac{n-\ell}n)^{n-\ell}]^{1-\frac 1n}\le 1\,.
$$
Now the LHS can be rewritten as $\frac{n}{\ell^{\frac \ell n}(n-\ell)^{\frac{n-\ell}n}}U$ where
$$
U={n-1\choose \ell-1}(\tfrac{\ell}n)^{\ell-1}(\tfrac{n-\ell}n)^{n-\ell}={n-1\choose \ell}(\tfrac{\ell}n)^{\ell}(\tfrac{n-\ell}n)^{n-\ell-1}
$$
Since $U$ appears as two equal terms in the binomial expansion of
$
[\tfrac{\ell}{n}+\tfrac{n-\ell}n]^{n-1}\,,
$
we must have $U\le \frac 12$. On the other hand, $\frac 1{\alpha^\alpha\beta^\beta}\le 2$ for $\alpha,\beta>0, \alpha+\beta=1$, and we are done again.
It would be interesting to see what factor on the right hand side can be put in place of $1$ for $n\ge 3$. I suspect that the worst case for even $n$ is addition of $n$ Rademacher independent random variables ($\pm 1$ with probability $\frac 12$ each) even if we allow them to be different (but with expectations of the same sign) but I have no proof of it at the moment. The odd case may have a rather ugly answer even for $n=3$.
A: A more conveniently written and slightly more general version of the desired result is as follows:
$$E\Big|\sum_{i\in[n]}X_i\Big|\ge E|X_1|,$$
where the $X_i$'s are iid random variables (r.v.'s) with a finite mean.
Let $p:=P(X_i\ge0)$, $q:=1-p$, $A:=EX_i^+$, $B:=EX_i^-$.
Without loss of generality, $0<p<1$.
For $J\subseteq[n]$, let
$$I_J:=1(X_j\ge0\ \forall j\in J,X_j<0\ \forall j\notin J).$$
Following the lines of the first part of the proof for $n=2$, we have
$$\begin{aligned}
E\Big|\sum_{i\in[n]}X_i\Big| 
&=\sum_{J\subseteq[n]}E\Big|\sum_{i\in[n]}X_i\Big|\,I_J \\ 
&=\sum_{J\subseteq[n]}E\Big|\sum_{i\in[n]}X_i\,
I_j\Big| \\ 
&=\sum_{J\subseteq[n]}E\Big|\Big(\sum_{i\in J}X_i^+-\sum_{i\notin J}X_i^-\Big)\,
I_J\Big| \\ 
&\ge\sum_{J\subseteq[n]}\Big|E\Big(\sum_{i\in J}X_i^+-\sum_{i\notin J}X_i^-\Big)\,
I_j\Big| \\ 
&=\sum_{J\subseteq[n]}\big||J|A/p-(n-|J|)B/q\big|\,p^{|J|}q^{n-|J|} \\ 
&=\sum_{k=0}^n\big|kA/p-(n-k)B/q\big|\,p^{k}q^{n-k}\binom nk=:s_n(p,A,B). 
\end{aligned}$$
Note also that
$$s_n(p,A,B)=E\big|AX/p-(n-X)B/q\big|,$$
where $X$ is a r.v. with the binomial distribution with parameters $n,p$.
Without loss of generality, $A+B=1$. It remains to show that
$$s_n(p,A,1-A)\ge1$$
for all $A\in[0,1]$. Graphing suggests this is true.
Unfortunately, the second, "squaring" part of the proof for $n=2$ will not work even for $n=3$ and $p$ close enough to $1/2$; that is, the inequality
$$\sum_{k=0}^n\Big(\big(kA/p-(n-k)B/q\big)\,p^{k}q^{n-k}\binom nk\Big)^2\ge(A+B)^2$$
will not hold for such $n,p$.
