Step 1: Recognise the object.
The quantity $\langle x \rangle = x - [x]$ is the fractional part of $x$. It always lies in $0 \le \langle x \rangle < 1$. We must decide which inequality between $\langle x+y \rangle$ and $\langle x \rangle + \langle y \rangle$ holds for every real pair.
Step 2: Split each number into integer plus fractional part.
Write $x = [x] + \langle x \rangle$ and $y = [y] + \langle y \rangle$. Adding, \[ x + y = \big([x] + [y]\big) + \big(\langle x \rangle + \langle y \rangle\big) \] The first bracket is an integer, so the fractional part of $x+y$ comes only from the second bracket.
Step 3: Look at the size of $\langle x \rangle + \langle y \rangle$.
Each fractional part is in $[0,1)$, so their sum lies in $[0,2)$. The behaviour splits at whether this sum crosses 1.
Step 4: Case where the sum stays below 1.
If $\langle x \rangle + \langle y \rangle < 1$, it is itself a valid fractional part, so \[ \langle x + y \rangle = \langle x \rangle + \langle y \rangle \] Here the two sides are equal.
Step 5: Case where the sum reaches or passes 1.
If $\langle x \rangle + \langle y \rangle \ge 1$, one whole unit is absorbed into the integer part, so \[ \langle x + y \rangle = \langle x \rangle + \langle y \rangle - 1 \] Since we subtract 1, the left side is strictly smaller.
Step 6: Combine both cases into one inequality.
In every case the left side is either equal to or one less than the right side, never greater. Therefore for all real $x,y$ \[ \langle x + y \rangle \le \langle x \rangle + \langle y \rangle \] which is exactly option (A).
\[ \boxed{\langle x+y \rangle \le \langle x \rangle + \langle y \rangle} \]