= Solution
The shuffle bound gives $\alpha\leq\beta\mathbin\#\beta$. By part (d), write
$$
\beta=\omega^\delta n+\eta,\qquad 1\leq n<\omega,\qquad\eta<\omega^\delta.
$$
Natural addition doubles corresponding Cantor coefficients, so
$$
\beta\mathbin\#\beta=\omega^\delta(2n)+(\eta\mathbin\#\eta)
<\omega^\delta(2n+1).
$$
On the other hand, ordinary ordinal addition absorbs each intermediate lower-order remainder, giving
$$
\beta+\beta+\beta=\omega^\delta(3n)+\eta.
$$
Consequently $\alpha<\beta+\beta+\beta$.
The stronger inequality with two copies is false. Take $\beta=\omega$, $\alpha=\omega+\omega$, let $X$ be the first block and $Y$ the second. Both have order type $\omega$, but $\alpha=\beta+\beta$.
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