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An integer-valued problem size requires integer subproblem sizes and a base case. For the displayed unequal split with $n\ge 2$, both children have positive size less than $n$, and $1/4\le m/n\le 1/2$. The entropy potential for a divide-and-conquer recurrence therefore supplies a uniform positive potential decrease. The squared sizes satisfy $m^2+(n-m{)}^2\le 5n^2/8$, which similarly closes a quadratic inductive hypothesis when the work at the root is $n^2$.

The printed fractional arguments require a size convention. To make the recurrence relation precise on positive integers, take a bounded base case $T(1)=t>0$ and use $m=\lceil n/4\rceil$, $n-m$ for $n\ge 2$. This integer rounding in a divide-and-conquer recurrence gives $1/4\le m/n\le 1/2$. The unrounded calculation is the special split fraction $1/4$.

Prove $T(n)\le An{\log }_{2}n+Bn$ by strong induction. Put $h_0=H_2(1/4)>0$, choose $A\ge 1/h_0$, and choose $B\ge t$. The base case follows. The entropy potential for a divide-and-conquer recurrence givesThe last step uses $H_2(m/n)\ge h_0$. This closes the inductive hypothesis and proves $\boxed{T(n)=O(n\log n)}$ in Big O notation.

Prove a proposed bound by mathematical induction, substituting the inductive hypothesis for smaller arguments into the recurrence relation. Choose its constants so the resulting inequality implies the proposed bound at the current argument, and verify all base cases. A leading-order guess may need an additional lower-order term to make the induction close. The entropy potential for a divide-and-conquer recurrence is an example using a linear correction to an $n\log n$ bound.