$\newcommand\abs[1]{\lvert#1\rvert}\newcommand\Abs[1]{\left\lvert#1\right\rvert}$**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, $2(A^2 + B^2) \ge (A + B)^2 = \left(\int\_0^1 \abs{f(x)} \,dx\right)^2$. $\square$