Integral inequality for continuous function Let $ f $ be a continuous, real-valued function on $[0, 1] $. 

Show that
  $$\int_0^1 \int_0^1 |f (x)+f (y)| dx dy \ge \int_0^1 |f (x)| dx $$

I tried to dissect the square in triangles and use some symmetry but it didn't work.
Can you help?
 A: Let us decompose $[0,1]^2$ as suggested by Martín-Blas by writing $P=\{x:f(x)>0\}$, $N=\{x:f(x)\le 0\}$ and write $g(x,y)=|f(x)+f(y)|$. Denote the Lebesgue measure on $[0,1]$ by $\lambda$. Then
$$\begin{align}\int_0^1 \int_0^1 |f(x)+f(y)| dx dy &= \int_{P\times P} g + \int_{P\times N} g + \overbrace{\int_{N\times P} g}^{=\int_{P\times N} g} + \int_{N\times N} g\\
&= 2\lambda(P) \int_{P} |f|+2\int_{P\times N} g + 2\lambda(N)\int_N |f|\tag{1}\\
\end{align}$$
Using the triangle inequality for integrals we get
$$\int_{P\times N} g\ge \left|\int_{P\times N} f(x)+f(y) d(x,y)\right|\tag{2}
=\left| \lambda(N) \int_P |f| - \lambda(P) \int_N |f|\right|
$$
Combining $(1)$ and $(2)$ we obtain
$$\begin{align}
\int_0^1 \int_0^1 |f(x)+g(x)| dx dy &\ge 2\lambda(P)\int_P |f| + 2\left| \lambda(N) \int_P |f| - \lambda(P) \int_N |f|\right|+2\lambda(N)\int_N |f|\tag{3}
\end{align}
$$
We may assume without loss of generality that $\lambda(P)\ge \lambda(N)$, because if the inequality that we want to prove holds for some $f$, then it also holds for $-f$.
Now there are two cases.
Case 1: $\int_P |f|\ge \int_N |f|$. Use $(3)$, drop the absolute values around the middle term and remember that $\lambda(P)+\lambda(N)=1$ to conclude that $(3)$ is greater-equal
$$\int_P |f|+\int_N |f|=\int_0^1 f(x) dx$$
Case 2: $\int_N |f|>\int_P |f|$. Then we also have $\lambda(P)\int_N |f|\ge \lambda(N)\int_P |f|$. That allows us to evaluate the absolute value in the middle term of $(3)$ and again conclude as in Case 1.
A: If $f(x)=0$ in $[0,1]$, the inequality is trivial to prove.
Now let $f(x)$ is not constant be zero in $[0,1]$. Consider the auxiliary function
\begin{equation}
F(\alpha)=\int_0^1\int_0^1\sqrt{f(x)^2+2\alpha f(x)f(y)+f(y)^2}dxdy
\end{equation}
Note that $F(1)=\int_0^1\int_0^1|f(x)+f(y)|dxdy$ and $F(0)\geq\int_0^1|f(x)|dx$. So what we going to prove is:
\begin{equation}
F(1)\geq F(0)
\end{equation}
Since $F(\alpha)$ is continuous about $\alpha$, from simple compute, we have:
\begin{equation}
F'(\alpha)=\int_0^1\int_0^1\frac{f(x)f(y)}{\sqrt{f(x)^2+2\alpha f(x)f(y)+f(y)^2}}dxdy
\end{equation}
Since $f(x)$ is not constant be zero in $[0,1]$, there must be $\inf\{f(x)^2+2\alpha f(x)f(y)+f(y)^2\}>0$ for $\alpha\in(0,1)$. What is more, if we assume $f(x)$ is bounded in $[0,1]$ (which is reasonable), then the $C=\sup\{f(x)^2+2\alpha f(x)f(y)+f(y)^2\}>0$ is existence. Thus, we have:
\begin{equation}
F'(\alpha)\geq C\int_0^1\int_0^1f(x)f(y)dxdy=C\int_0^1f(x)dx\int_0^1f(y)dy=C\left(\int_0^1f(x)dx\right)^2>0
\end{equation}
Thus, we can claim $F(0)\leq F(1)$ which is going to prove.
A: Try this: $P=\{x\in[0,1]\,\vert\,f(x)\ge 0\}$, $N=\{x\in[0,1]\,\vert\,f(x)<0\}$ and decompose the domain:
$$[0,1]\times[0,1]=(P\times P)\cup(P\times N)\cup(N\times P)\cup(N\times N).$$
A: lemma: let $x_{i}\in R$,then we have
$$\sum_{1\le i,j\le n}|x_{i}+x_{j}|\ge n\sum_{k=1}^{n}|x_{k}|$$
This post have solution:How prove this inequality: $\sum_{i,j=1}^{n}|x_{i}+x_{j}|\ge n\sum_{i=1}^{n}|x_{i}|$?
then let $x_{i}\to f(x_{i}),x_{j}\to f(x_{j})$
$$\Longrightarrow \dfrac{1}{n^2}\sum_{i,j=1}^{n}|f(x_{i})+f(x_{j})|\ge\dfrac{1}{n}\sum_{i=1}^{n}|f(x_{i})|$$
$$\Longrightarrow \dfrac{1}{n^2}(n^2-n)\int_{0}^{1}\int_{0}^{1}|f(x)+f(y)|dxdy+\dfrac{2n}{n^2}\int_{0}^{1}|f(x)|dx\ge\dfrac{n}{n}\int_{0}^{1}|f(x)|dx$$
then we have
$$\int_{0}^{1}\int_{0}^{1}|f(x)+f(y)|dxdy\ge\dfrac{n-2}{n-1}\int_{0}^{1}|f(x)|dx$$
let $n\to \infty$,then
$$\int_{0}^{1}\int_{0}^{1}|f(x)+f(y)|dxdy\ge\int_{0}^{1}|f(x)|dx$$
