This is inspired by an old Putnam problem from 2003, and a solution given by composite of solutions by Kent Merryfield:

Question (Putnam 2003)Let $f(x)$ be a continuous real-valued function defined on the interval $[0, 1]$. Show that $$\int_0^1\int_0^1|f(x)+f(y)|dx \; dy \ge \int_0^1 |f(x)|dx$$ Kent Merryfield's proof:

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 |f(x) + f(y)|\,dx\,dy &= \int_P \int_P |f(x) + f(y)|\,dx,dy + \int_P \int_N |f(x) + f(y)| \,dx\, dy \\ &+ \int_N \int_P |f(x) + f(y)|\, dx \,dy+ \int_N \int_N |f(x) + f(y)|\, dx\, dy\end{align*} We tackle these terms one at a time. \begin{align*}\int_P \int_P |f(x) + f(y)|\, dx\, dy &= \int_P \int_P (f(x) + f(y))\,dx \,dy\\ &= |P| \int_P f(x)\,dx + |P| \int_P f(y) \,dy = 2|P| \int_P f^+(x) \,dx\end{align*} where we use the notation $|P|$ to mean the measure (total net length) of the set $P.$

Similarly, $\int_N \int_N |f(x) + f(y)| \,dx \,dy = 2|N| \int_N f^-(x) \,dx.$

The other two terms are equal to each other (as shown by interchanging $x$ and $y$). \begin{align*}\int_P \int_N |f(x) + f(y)| \,dx \,dy &= \int_P \int_N |f^+(x) - f^-(y)| \,dx \,dy\\ &\ge \left|\int_P \int_N f^+(x) - f^-(y) \,dx \,dy\right|\\ &= \left| |N| \int_P f^+(x) \,dx - |P| \int_N f^-(y) \,dy \right|\end{align*} If we let $A = \int_P f^+(x) \,dx,$ $B = \int_N f^-(x) \,dx,$ and $I = \int_0^1 \int_0^1 |f(x) + f(y)| \,dx \,dy,$ then we have found that $I \ge 2|P|A + 2|N|B + 2|(|N|A - |P|B)|.$

For convenience, we now square this: \begin{align*}I^2 &\ge 4\left[(|P|A + |N|B)^2 + (|N|A - |P|B)^2 + (\text{other positive terms})\right]\\ &\ge 4(|P|^2A^2 + |N|^2B^2 + |N|^2A^2 + |P|^2B^2)\\ &= 4(|P|^2 + |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, $2(|P|^2 + |N|^2) \ge (|P| + |N|)^2 = 1^2,$ since it is the measure of the interval $[0, 1].$ Also, $2(A^2 + B^2) \ge (A + B)^2 = \left(\int_0^1 |f(x)| \,dx\right)^2 .$

MY Question 1:Let $f(x)$ be a continuous real-valued function defined on the interval $[0, 1]$. Show that $$\int_{0}^{1}\int_0^1\int_0^1|f(x)+f(y)+f(z)|dx \; dy\;dz \ge \int_0^1 |f(x)|dx$$ and I think maybe this generality also is hold:

MY Question 2:Let $f(x)$ be a continuous real-valued function defined on the interval $[0, 1]$. Show that $$\int_{0}^{1}\int_{0}^{1}\cdots\int_0^1\int_0^1|f(x_{1})+f(x_{2})+\cdots+f(x_{n})|dx \; dx_{2}\cdots;dx_{n} \ge \int_0^1 |f(x)|dx$$ and Now How to prove.Thanks