Let $k\geq 2$ and 
$$f(x_1,\ldots,x_k)=\Bigl(\prod_{i\leq k}(1+x_i)+\prod_{i\leq k}(1-x_i)\Bigr)\log\Bigl(\prod_{i\leq k}(1+x_i)+\prod_{i\leq k}(1-x_i)\Bigr)-k(1-x_1 x_2)\log(1-x_1 x_2).$$
If random variables $X_1,\ldots, X_k \in (-1,1)$ are i.i.d. with $\mathbb{E}X_1=0$ and $\varepsilon_1,\ldots,\varepsilon_k$ are i.i.d. random signs (symmetric, and independent of $X_1,\ldots, X_k$), is it true that
$$\mathbb{E}f(X_1,\ldots,X_k)\leq \mathbb{E}f(\varepsilon_1 X_1,\ldots,\varepsilon_k X_k)?$$
For $k=2$ this follows by Taylor's expansion, and numerical simulations seem to support this for $k\geq 3$.