Let $f:\left\{-1,1\right\}^n\to\left\{-1,1\right\}$.
How to show that $$ P_{{x,y,z}} \{f(xyz)=f(x)f(y)f(z)\} \ge 1/2? $$ where $x,y,z$ are distributed uniformly and independently on $\left\{-1,1\right\}^n$.
Equivalently, the set $$ \{(x,y,z)\in (\left\{-1,1\right\}^n)^3 \,|\,f(xyz)=f(x)f(y)f(z)\} $$ has at least $2^n\times 2^n\times 2^{n-1}$ elements.
I think there should be a simple argument that I am missing.
Edit:
I forgot to define: $$ (xy)_i:=x_iy_i, \,\,\, (xyz)_i:=x_iy_iz_i, $$ so the product is component-wise.
I will reformulate.
Take the space $\mathbb{F}_2^n$ and define the function $u:\mathbb{F}_2^n\rightarrow \mathbb{F}_2$ via $$f(z)=(-1)^{u(x)}$$ where $z=(z_1,\ldots,z_n)\in \{-1,+1\}^n$ and $x=(x_1,\ldots,x_n)\in \mathbb{F}_2^n.$ Since all $\mathbb{F}_2^n$ operations are modulo $2,$ and the products of $f$ become sums of $u$ what we desire to show is $$ \sum_{x,y,z}(-1)^{u(x)+u(y)+u(z)+u(x+y+z)}\geq 0,\qquad(1) $$ where the sums are over $\mathbb{F}_2^n.$ This is because $$ f(xyz)=f(x)f(y)f(z) \Leftrightarrow u(x+y+z)=u(x)+u(y)+u(z), $$ and when the equality holds we pick up a $+1$ in the sum, otherwise we pick up a $-1.$ Now the sum in $(1)$ can be rewritten as $$ \sum_{x,y}(-1)^{u(x)+u(y)}\sum_{z}(-1)^{u(z)+u(x+y+z)}. $$ For each fixed $x,y$ letting $a=x+y,$ the inner sum is of the form $$ \sum_{z}(-1)^{u(z)+u(a+z)}, $$ and when $a=0$ which is the same as $x=y,$ the inner sum takes on the value $2^n$ since the exponent is $2u(z)$ which is zero modulo $2.$ This happens for $2^n$ pairs $(x,y)$ with $y=x.$
When $a\neq 0,$ then we need to treat the sum differently. Let us rewrite the whole sum as $$ \sum_{x,y:x=y}(-1)^{2u(x)}\sum_{z}(-1)^{2u(z)}+ \sum_{x,y:x\neq y}(-1)^{u(x)+u(y)}\sum_{z}(-1)^{u(z)+u(x+y+z)}, $$ which becomes $$ 2^{2n}+\sum_{x}\sum_{a\neq 0}(-1)^{u(x)+u(x+a)}\sum_{z}(-1)^{u(z)+u(a+z)}, $$ or $$ 2^{2n}+\sum_{a\neq 0}\sum_{x}(-1)^{u(x)+u(x+a)}\sum_{z}(-1)^{u(z)+u(a+z)}, $$ or $$ 2^{2n}+\sum_{a\neq 0}\left(\sum_{x}(-1)^{u(x)+u(x+a)}\right)^2\geq 0, $$ as required. When $u$ is linear, i.e., $$u(x+a)=u(x)+u(a),\quad \forall x,a$$ the second term is actually zero since a linear function is balanced.