Ternary lower-bound search

A lower-bound search that probes two positions to retain one of three subintervals. For unresolved $[lo,hi)$ of positive length $d$, put $q=\lfloor d/3\rfloor$, $m_1=lo+q$ and $m_2=hi-1-q$. If $a(m_1)\ge x$, retain $[lo,m_1)$; if $a(m_2)<x$, retain $[m_2+1,hi)$; otherwise retain $[m_1+1,m_2)$. Sortedness preserves the lower-bound partition. The new interval lengths are $q,q,d-2q-2$, and the feasible middle case has nonnegative length at most $q$. This proves strict progress and $O(\log (N+1))$ comparison cost. Equality must retain the first possible boundary rather than return an arbitrary matching occurrence.