How to prove this inequality? For any nonnegative random variable $X$ independent of $U$ where $U \sim \operatorname{Uniform}(-t,t)$ and any $t\ge 0,$
$$P(X+U\ge t)\le\frac{E(X)}{2t}.$$
Any hints to prove this inequality?

Yes, your comment is on the right track.  This looks like it's going to be Markov's Inequality:
$$P(Z\geq z)\leq \frac{E[Z]}{z}$$
for non-negative $Z$. As you have noted, $X+U+t$ is non-negative, so it's a candidate for $Z$, and $X+U+t\geq 2t$ when $X+U\geq t$. So, consider the mean of $Z$.
