$\int_0^1\int_0^1|f(x)+f(y)|\mathrm{d}x\mathrm{d}y\ge\int_0^1|f(x)|\mathrm{d}x$ Suppose $f(x)$ is integrable on $[0,1]$. Prove:
$$\int_0^1\int_0^1|f(x)+f(y)|\mathrm{d}x\mathrm{d}y\ge\int_0^1|f(x)|\mathrm{d}x$$
I think I proved it in its discrete form. (It seems like an Olympic problem)
Is there any better method? Thanks for any help!
 A: In probabilistic terms this inequality says the following: If $X_1$ and $X_2$ are two independent realizations of the real random variable $X$ then
$$E\bigl(|X_1+X_2|\bigr)\geq E\bigl(|X|)\ .\tag{1}$$ If $X$ takes only  values $\geq0$ this is obvious (even with a factor $2$ on the right hand side), but otherwise one might think that there could be so much cancellation that $E\bigl(|X_1+X_2|\bigr)$ is in fact smaller than $E\bigl(|X|)$. I guess that $(1)$ is well known in probability circles.
It is enough to prove the stated inequality for a function $f$ that takes only finitely many values $f_i\in{\mathbb R}$ on intervals of length $p_i>0$.  We may assume that ${\rm sgn}(f_i)={\rm sgn}(i)$, and that
$$p:=\sum_{i<0} p_i>0, \quad a:={1\over p}\sum_{i<0} |f_i| p_i>0,\quad q:=\sum_{i\geq0}p_i>0,\quad b:={1\over q}\sum_{i\geq0} f_ip_i>0\ .$$
The quantity
$$U:=\int_{[0,1]^2}\bigl|f(x)+f(y)\bigr|\>{\rm d}(x,y)=\sum_{i, \>k}|f_i+f_k|\>p_i p_k$$
then is a sum of four partial sums that can be dealt with as follows:
$$\sum_{i\geq0, \>k\geq0}|f_i+f_k|\>p_i p_k=\sum_{i\geq0, \>k\geq0}\bigl(f_i+f_k\bigr)\>p_i p_k=2q^2b\ ,$$
and similarly
$$\sum_{i<0, \>k<0}|f_i+f_k|\>p_i p_k=\sum_{i<0, \>k<0}\bigl(|f_i|+|f_k|\bigr)\>p_i p_k=2p^2a\ .$$
For the "unequal sign" sums we make use of the convexity of the functions $\phi_c(x):=|x+c|$. This allows to write
$$\sum_{i\geq0, \>k<0}|f_i+f_k|\>p_i p_k=\sum_{i<0, \>k\geq0}|f_i+f_k|\>p_i p_k\geq p\sum_{k\geq0}|{-a}+f_k| p_k\geq pq|b-a|\ .$$
In all we have proven that
$$U\geq 2q^2 b+2p^2 a+2pq|b-a|\ .$$
On the other hand
$$V:=\int_0^1 |f(x)|dx=\sum_i|f_i| \>p_i=pa+qb \ .$$
By symmetry and linearity we may assume $b\geq a$ and $a=1$, hence $b=1+h$, $h\geq0$. As $q=1-p$ we therefore have to prove that
$$U-V\geq 2(1-p)^2(1+h)+2p^2+2p(1-p)h-\bigl(p+(1-p)(1+h)\bigr)\tag{2}$$ is $\geq0$ for $0\leq p\leq1$. The function
$$\psi(p):=1+h-(4+h)p+4p^2$$
appearing on the right hand side of $(2)$ takes its global minimum on ${\mathbb R}$ at the point $p_*={1\over2}+{h\over8}$. If $h\geq4$ then $p_*\geq1$, hence $\psi$ is decreasing for $0\leq p\leq1$. It follows that in this case $$U-V\geq\psi(1)=1>0\ .$$ If $0\leq h<4$ then
$$U-V\geq \psi(p_*)={h\over2}\left(1-{h\over8}\right)\geq0\ .$$
In particular we have equality in $(1)$ if $X$ assumes the values $\pm1$ with probability ${1\over2}$ each.
