Let $X \in [m,M]$ be a random variable and $S$ be the Rademacher random variable (e.i $P(S=1)=P(S=-1)=1/2$). Is the following inequality true?
$$ P(X-EX \geq t) \leq P((M-m)S \geq 2t) $$
This inequality showed up while I was trying to prove the Hoeffding's inequality, and I was wondering if someone could help me either prove or disprove it.
The more general inequality is the following. For $X_i \in [m_i,M_i]$ independent, and $S_i\sim Rademacher$ also independent, one should prove that
$$ P(\sum^n_{i=1}X_i - EX_i \geq t) \leq P(\sum^n_{i=1}(M_i - m_i)S_i \geq 2t) $$
I think it is not true.
It is quite well possible to create a random variable $X\in\left[-\frac{1}{2},\frac12\right]$ that is distributed in such a way that $\mathbb{E}X<0$ and $P\left(X< x\right)<1$ for every $x<\frac12$.
If the inequality is correct then for any $t>\frac{1}{2}$ we find: $$P\left(X-\mathbb{E}X\geq t\right)\leq P\left(S\geq2t\right)=0$$
This implies that: $$P\left(X<\mathbb{E}X+\frac{1}{2}\right)=1$$ However this contradicts that $P\left(X< x\right)<1$ for every $x<\frac12$.