I am trying to show the inequality below, please note that in this case I am considering the labels to be $Y_i \in \{0, 1\}$, I state this since I have seen results but for labels that are in $\{0,1\}$. The inequality I want to show is:
$$ \mathbb{E}\max_{f\in F}\left[ \frac{1}{n}\sum_{i=1}^n\epsilon_i\mathbb{1} \{ y_i\neq f(x_i) \} \right]\leq \mathbb{E}\max_{f\in F}\left[ \frac{1}{n}\sum_{i=1}^n\epsilon_i f(x_i) \right] $$
where $\epsilon_i$ is a Rademacher random variable and the labels $Y_i\in \{ 0,1 \}$ as well as the outputs of $f$, the $\mathbb{1} \{ y_i\neq f(x_i) \}$ is the indicator function.
My attempt of solution: I rewritten the indicator function as follows: $$ \mathbb{1} \{ y_i\neq f(x_i)\}=y_i + f(x_i) - 2y_if(x_i) $$ And replaced it into the expression that became the following:
$$ \mathbb{E}\max_{f\in F}\left[\frac{1}{n}\sum_{i=1}^n\epsilon_i\mathbb{1} \{ y_i\neq f(x_i) \} \right]=\mathbb{E}\max_{f\in F}\left[\frac{1}{n}\sum_{i=1}^n\epsilon_i f(x_i)-\frac{1}{n}\sum_{i=1}^n\epsilon_i (2f(x_i)-1)y_i \right] $$
I rewritten it this way since the term $(2f(x_i)-1)$ has a similar distribution as the Rademacher random variable, but I don't know how to continue or if this approach is correct, thanks for any help.