For $X_i \sim$ i.i.d with cdf $F_x$, and $\forall c \in \mathbb R$, then, letting $M_n$ denote the maximum observation
$$ M_n \le c+ \sum_i^n (X_i - c) \mathbb I(X_i > c) $$ I proved this by taking $n=1$ and $X_i = c$ and showing that this was a contradiction. Now, I am trying to show that, for $\bar{F}(x)= 1- F(x)$ :
$$ E(M_n) \le c + n \int_c^\infty \bar{F}(x)dx $$ My attempt so far, using the initial result, then then the same inequality must hold for the expectations of both sides, that is: \begin{align*} E(M_n) &\le c+ \sum_i^n E[(X_i - c) \mathbb I(X_i > c)] \\ &\overset{\text{iid}}{=}c+ n E[(X - c) \mathbb I(X > c)] \\ &= c+ n \int_{-\infty}^ \infty (x-c)\mathbb I(X > c) f_X(x) dx \\ &= c+ n \int_{c}^ \infty (x-c) f_X(x) dx \\ &= c+ n \int_{c}^ \infty x f_X(x) dx - nc \int_c^\infty f_X(x)dx \\ &=c+ n \int_{c}^ \infty x f_X(x) dx -nc \bar{F}(c)\\ \end{align*}
I think i take a wrong turn somewhere, maybe I shouldn't expand the term in the integral, I still don't quite get where the second integral would come from, to define the CDF in the required inequality... any hints?