Prop. Let $f\colon [0,1]\times[0,1] \to [0,1]$ be a symmetric measurable function. Then we have $$ \iiint\limits_{[0,1]^3} \prod_{\mathrm{cycl}}f(x,y) \mathrm{d}x\mathrm{d}y\mathrm{d}z +\iiint\limits_{[0,1]^3} \prod_{\mathrm{cycl}}(1-f(x,y))\mathrm{d}x\mathrm{d}y\mathrm{d}z \geq \frac{1}{4} $$
After expanding a bit, the $f(x,y)f(y,z)f(z,x)$ term cancels, and what remains is the following claim:
$$ \iiint\limits_{[0,1]^3} (1-f(x,y)-f(y,z)-f(z,x) +f(x,y)f(y,z)+f(y,z)f(z,x)+f(z,x)f(xy)) \mathrm{d}x\mathrm{d}y\mathrm{d}z \geq \frac{1}{4} $$
By linearity, we get:
$$ 1-3\iint\limits_{[0,1]^2}f(x,y)\mathrm{d}x\mathrm{d}y+3\iiint\limits_{[0,1]^3}f(x,y)f(y,z) \mathrm{d}x\mathrm{d}y\mathrm{d}z \geq \frac{1}{4}$$
or
$$ \frac{1}{4}+\iiint\limits_{[0,1]^3}f(x,y)f(y,z) \mathrm{d}x\mathrm{d}y\mathrm{d}z \geq \iint\limits_{[0,1]^2}f(x,y)\mathrm{d}x\mathrm{d}y $$
or
$$ \iiint\limits_{[0,1]^3}f(x,y)(1-f(y,z))\mathrm{d}x\mathrm{d}y\mathrm{d}z \leq \frac{1}{4} $$
Any ideas on how to prove this one?
We have: $$\begin{align} I&:=\iiint\limits_{[0,1]^3} (1-\underbrace{(f(x,y)+f(y,z)+f(z,x))}_{\text{symmetry} \Longrightarrow3f(x,z)} +\underbrace{(f(x,y)f(y,z)+f(y,z)f(z,x)+f(z,x)f(xy))}_{\text{symmetry} \Longrightarrow3f(z,x)f(y,z)}) \mathrm{d}x\mathrm{d}y\mathrm{d}z \\ &=\iiint\limits_{[0,1]^3} (1-3f(x,y)+3f(x,y)f(y,z)) \mathrm{d}x\mathrm{d}y\mathrm{d}z \hspace{1cm} \text{symetry of }(x,y,z) \text{ over }[0,1]^3\\ &=\iiint\limits_{[0,1]^3} 1dxdydz -3\iiint\limits_{[0,1]^3} f(x,y) \mathrm{d}x\mathrm{d}y\mathrm{d}z +3\iiint\limits_{[0,1]^3} (f(x,y)f(y,z) \mathrm{d}x\mathrm{d}y\mathrm{d}z \end{align}$$ The first integral is equal to $$I_1:=\int\limits_{[0,1]} 1dy$$ For the second integral: $$I_2:=\iiint\limits_{[0,1]^3} f(x,y) \mathrm{d}x\mathrm{d}y\mathrm{d}z=\iint\limits_{[0,1]^2} f(x,y) \mathrm{d}x\mathrm{d}y = \int_{[0,1]}\left(\int\limits_{[0,1]} f(x,y)\mathrm{d}x\right)dy$$ For the second integral, as the function $f$ is symmetric $$\begin{align} I_3 &:=\iiint\limits_{[0,1]^3} f(x,y)f(y,z) \mathrm{d}x\mathrm{d}y\mathrm{d}z \\ &=\int_{[0,1]}\left(\iint\limits_{[0,1]^2} f(x,y)f(z,y) \mathrm{d}x\mathrm{d}z\right)dy \hspace{0.5cm}\text{as the function $f$ is symmetric, $f(y,z)=f(z,y)$}\\ &=\int_{[0,1]}\left(\int\limits_{0\le z\le 1}f(z,y)\left(\int_{0\le x\le 1} f(x,y)dx \right) \mathrm{d}z\right)dy\\ &=\int_{[0,1]}\left(\left(\int_{0\le x\le 1} f(x,y)dx\right)\left(\int_{0\le z\le 1}f(z,y) dz\right)\right)dy\\ &=\int_{[0,1]}\left(\int\limits_{[0,1]^2} f(x,y)\mathrm{d}x\right)^2dy \end{align}$$
Finally, $$\begin{align} I &= I_1 - 3I_1 +3I_2 \\ &= \int_{[0,1]}\left(1 -3\left(\int\limits_{[0,1]^2} f(x,y)\mathrm{d}x\right) +3\left(\int\limits_{[0,1]^2} f(x,y)\mathrm{d}x\right)^2\right) dy\\ &= \int_{[0,1]}\left(\frac{1}{4} +3\left(-\frac{1}{2}+\int\limits_{[0,1]^2} f(x,y)\mathrm{d}x\right)^2\right) dy \ge \int_{[0,1]}\left(\frac{1}{4} \right) = \frac{1}{4} dy \end{align}$$ The equality occurs iff the function $f(x,y)$ satisfies $$\int_{0\le x\le 1}f(x,y)dx = \frac{1}{2} \hspace{1cm} \text{for all $y\in [0,1]$}$$