Asymptotic of divide-and-conquer type recurrence with non-constant weight repartition between subproblems and lower order fudge terms
According to the OP, to complete the proof we need to prove that for large enough $n$,
$$
(a_n n + \sqrt{n})^{a_n + 1/\sqrt{n}}((1-a_n)n + \sqrt{n})^{1-a_n + 1/\sqrt{n}} \leq n.
$$
Taking out a factor of $n^{1+2/\sqrt{n}}$, we get
$$
(a_n + 1/\sqrt{n})^{a_n + 1/\sqrt{n}}
(1-a_n + 1/\sqrt{n})^{1-a_n + 1/\sqrt{n}} \leq n^{-2/\sqrt{n}}.
$$
Dividing by $(1+2/\sqrt{n})^{1+2/\sqrt{n}}$, we get
$$
\left( \frac{a_n + 1/\sqrt{n}}{1 + 2/\sqrt{n}} \right)^{a_n + 1/\sqrt{n}}
\left( \frac{1-a_n + 1/\sqrt{n}}{1 + 2/\sqrt{n}} \right)^{1-a_n + 1/\sqrt{n}} \leq
\frac{1}{n^{2/\sqrt{n}} (1+2/\sqrt{n})^{1+2/\sqrt{n}}}.
$$
Raising to the power $1/(1+2/\sqrt{n})$, and defining $p = (a_n + 1/\sqrt{n})/(1 + 2/\sqrt{n})$, we get
$$
p^p (1-p)^{1-p} \leq \frac{1}{n^{(2/\sqrt{n})/(1+2/\sqrt{n})}(1+2/\sqrt{n})}.
$$

The left-hand side is $1/\exp h(p)$, where $h(p)$ is the entropy function. Hence it is maximized when $a_n = 1/4$, at which point $p \approx 1/4$. Since $(1/4)^{1/4} (3/4)^{3/4} &lt; 1$, for large enough $n$ we can bound the left-hand side by some $\theta &lt; 1$.
As for the right-hand side, as $n\to\infty$ it tends to $1$. Indeed, taking logarithm, we get
$$
\frac{2\log n}{\sqrt{n} + 2} + O\left(\frac1{\sqrt{n}}\right) \longrightarrow 0.
$$
In particular, for large enough $n$ it will be larger than $\theta$.