Non-comparable natural numbers
When you say "undecidable" I assume you mean it is independent of a theory such as ZFC. There will be statements like
$$B(m)&gt;n$$
(for natural numbers $m$, $n$) that are not decided by ZFC, assuming ZFC is consistent. Because otherwise we could compute the function $B$ just by searching for proofs in ZFC of such statements.

Since $B$ is Turing complete there is some Turing machine $\Phi$ with
Con(ZFC)$\iff$ $\Phi$ accepts with oracle $B$ (on input 0, say) and $\neg$Con(ZFC)$\iff$ $\Phi$ rejects.

Now assuming that actually Con(ZFC) is true we know $\Phi$ accepts and there is some collection of facts $B(m_i)=n_i$, $1\le i\le k$ that was used in the computation (we may set it up so that the oracle access works in this way). Then
$$\sum_{i=1}^k (B(m_i)-n_i)^2&gt;0$$
is false, but this fact is not provable in ZFC, else ZFC would prove its own consistency.
Of course this can be rewritten as
$$\sum_{i=1}^k B(m_i)^2+n_i^2&gt;\sum_{i=1}^k 2B(m_i)n_i\tag{*}$$
and so arguably (*) provides a Yes answer to your question.

However, I don't think we can figure out what these numbers $m_i$, $n_i$ are, because the queries are adaptive (what is asked about depends on answers to previous questions, and we don't know those answers).