Let $F\subset \mathbb{R}^n$ be a finite set and $\sigma$ be uniformly distributed over $\{-1,1\}^n$. The usual Rademacher average of $F$ (modulo normalizing factors) is $$ R_n(F)=\mathbb{E}_\sigma \max_{f\in F}\sum_{i=1}^n \sigma_if_i. $$ Now let us define two operations on $F$: $\mathrm{conv}(F)$ and $[F]_\vee$. The former is just the convex hull of the vector-set $F$ in $\mathbb{R}^n$. The latter is defined by $$ [F]_\vee=\{ f\vee g :f,g \in F\}, $$ where $(f\vee g)_i=\max\{f_i,g_i\}$ is the coordinate-wise maximum. It is well-known (and easy to show) that $ R_n( \mathrm{conv}(F)) = R_n( F) $.

Question: is it true that $$ R_n( [\mathrm{conv}(F)]_\vee) = R_n( [F]_\vee) $$ ?

The inequality $ R_n( [F]_\vee) \le R_n( [\mathrm{conv}(F)]_\vee) $ holds due to set containment. Also, $\mathrm{conv}(F)$ is an infinite set, so the $\max$ in the definition should be replaced by a $\sup$.

**Update**. Fedor Petrov has constructed a counterexample, which I've accepted. The more general conjecture I had was as follows. Define the $k$-fold max operator $[F]_k$ by
$$ [F]_k=\{ f_1\vee f_2\ldots\vee f_k :f_i \in F\}.
$$
Is there a universal constant $c$ (independent of $n$ and $k$) such that
$$
R_n( [\mathrm{conv}(F)]_k)
\le
c R_n( [F]_k)
$$?