Inequality involving $X_{(n)}$ I am trying to prove that for random variables (not necessarily iid) $X_1,...,X_n$ that $X_{(n)} \le x + \sum_{i=1}^n  X_i \mathbb{I}_{X_i > x}$ for all $x \in \mathbb R$.
I tried to show this by a contraction, which would lead to the assumption that:
$$
E (X_{(n)}) \overset{\text{assumed}}{>} x +\sum_{i=1}^n E[ X_i \mathbb{I}_{X_i > x}] \overset{\text{Markov Inequality}}{\ge} x+\sum_{i=1}^n  E[X_i]
$$
but I keep going around in circles, and since the rvs are not iid, it is confusing to work with the maximum, any hints?
update the inequality will only hold if $X_i$'s are assumed to be non-negative and $x \ge 0$
 A: This is not true if the $X_i$ can be negative :
take $n=1$, $x=-2$ and $X_1 = -1$. Then 
$$x+ X_{1}\Bbb 1_{X_{1}>x}= -2+(-1) = -3 < -2 = X_{(1)}$$
It is also not true if $x$ can be negative :
take $n=1$, $x=-1$ and $X_1 = 1$ Then 
$$x+ X_{1}\Bbb 1_{X_{1}>x}= -1+1 = 0 < 1 = X_{(1)}$$
Now, if all the $X_i$ are positive, and that $x\geq 0$, you have 
$$\forall \omega, X_{(n)}(\omega) = X_{(n)}(\omega) \Bbb 1_{X_{(n)} \leq x} (\omega) + X_{(n)} \Bbb 1_{X_{(n)} > x} (\omega) \leq x +  X_{(n)} \Bbb 1_{X_{(n)} > x} (\omega)$$
But $\forall \omega, \exists i, X_{(n)} (\omega) = X_i(\omega)$, so
$$ \forall \omega, X_{(n)} (\omega) \leq \sum_{i=1}^n X_i(\omega)$$
And you also have $$ \Bbb 1_{X_{(n)} > x} \leq \Bbb 1_{X_{i} > x} $$
So
$$ \forall \Bbb 1_{X_{(n)} > x}(\omega) X_{(n)} (\omega) \leq \sum_{i=1}^n X_i(\omega)1_{X_{(n)} > x}(\omega) \leq \sum_{i=1}^n X_i(\omega)1_{X_{i} > x}(\omega) $$
Hence the result
$$X_{(n)} \leq  x+ \sum_{i=1}^n X_i(\omega)1_{X_{i} > x}(\omega) $$
A: For $x \in \mathbb{R}_+$
\begin{align}
\max_{1\le i\le n}\{|X_i|\}&=\max_{1\le i\le n}\{|X_i|1\{|X_i|\le x\}+|X_i|1\{|X_i|> x\}\} \\[1.7ex]
&\le \max_{1\le i\le n}\{x+|X_i|1\{|X_i|> x\}\} \\
&=x+\max_{1\le i\le n}\{|X_i|1\{|X_i|> x\}\} \le x+\sum_{i=1}^n |X_i|1\{|X_i|> x\}.
\end{align}

If $X_i\ge 0, i=1,\dots,n$, then $X_i=|X_i|$ and
$$\max_{1\le i\le n}\{X_i\}\le x+\sum_{i=1}^n X_i1\{X_i> x\}$$.
