Maximal Probability inequality In lecture notes to a lecture I'm currently attending my teacher claims that the following holds true:

Let $X,Y_1,\dots,Y_n$ be random variables on the same probability space. Then for any $x > 0$ the following holds true $$\mathbb{P}\left(\max_{1\leq j \leq n}\vert X - Y_j\vert > x\right) \leq \mathbb{P}\left(\vert X\vert > x/2\right) + \mathbb{P}\left(\max_{1\leq j \leq n}\vert Y_j\vert > x/2\right).$$

I tried to verify this, but this seems not correct. What am I missing? Note that we assume neither independence nor equality in distribution. Do we maybe need such an assumption?
In my attempt to verify this, my approach was to reduce the problem to the case $n=1$. But even this didn't work to.
 A: The inclusion
$$\tag{*}
\left\{\max_{1\leqslant j\leqslant n}\lvert X-Y_j\rvert>x\right\}\subset
\{\lvert X\rvert >x/2\}\cup \left\{\max_{1\leqslant j\leqslant n}\lvert Y_j\rvert>x/2\right\}
$$
takes place. Indeed, if $\omega$ is in the right hand side of $(*)$, then $\lvert X(\omega)\rvert \leqslant x/2$ and $\max_{1\leqslant j\leqslant n}\lvert Y_j(\omega)\rvert \leqslant x/2$ hence
$$
\max_{1\leqslant j\leqslant n}\lvert X-Y_j\rvert\leqslant \max_{1\leqslant j\leqslant n}\lvert X(\omega)\rvert+\max_{1\leqslant j\leqslant n}\lvert Y_j(\omega)\rvert= \lvert X(\omega)\rvert+\max_{1\leqslant j\leqslant n}\lvert Y_j(\omega)\rvert\leqslant x
$$
which means that $\omega$ belongs to the complement of the left hand side of $(*)$. Now use the elementary fact that $A\subset B\cup C$ implies
$$
\mathbb P(A)\leqslant \mathbb P(B\cup C)\leqslant \mathbb P(B)+\mathbb P(C).
$$
A: Hint:
If $|X|\leq\frac{1}{2}x$ and $\max_{1\leq i\leq n}|Y_{i}|\leq\frac{1}{2}x$ or equivalently
$X,Y_{1},\dots,Y_{n}\in\left[-\frac{1}{2}x,\frac{1}{2}x\right]$ then
$|X-Y_{i}|\leq x$ for every $i\in\{1,\dots,n\}$ or equivalently $\max_{1\leq i\leq n}|X-Y_{i}|\leq x$.
So $\max_{1\leq i\leq n}|X-Y_{i}|>x$ implies that $|X|>\frac{1}{2}x$ or $\max_{1\leq i\leq n}|Y_{i}|>\frac{1}{2}x$
