= Solution
Put $m=|w|$ and $\ell=|r|$. In fact minimality yields the stronger bound $m\leq\lfloor\ell/2\rfloor$. Suppose instead that
$$
m\geq k:=\lfloor\ell/2\rfloor+1.
$$
Because $|u|>\ell/2$, the first $k$ letters of the matched segment $u$ form a <relator> segment $v$ of length $k$. They also form a segment of the periodic word $w^n$. Since $k\leq m$, that segment is contained in one cyclic permutation $w'$ of $w$, starting at the same position in the period. A cyclic permutation represents a conjugate and has length $m$.
Rotate the symmetrized <relator> to write it as $vs$. Replacing the prefix $v$ of $w'$ by $s^{-1}$ preserves the represented element and gives a word of length at most
$$
m-k+(\ell-k)=m+\ell-2k<m.
$$
<Free reduction> can only shorten further. This contradicts the choice of $w$. Thus
$$
\boxed{|w|\leq\lfloor|r|/2\rfloor\leq |r|/2+2.}
$$
The periodic-segment argument treats matches that cross a boundary between copies of $w$; it would be incorrect simply to assume that the entire long segment $u$ lies in the original written copy of $w$.
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