In this question, the asker specified without derivatives and I provided an answer using the most basic equivalents which gives me a quite accurate result.
Later, I made the problem more general $$n^{\sqrt{n+a}}>(n+b)^{\sqrt{n+b}}\implies n > \,\, ??? $$ and looked (as explained in my answer) for an estimate , in the real domain, of the zero of $$g(n)=\sqrt{n+a} \log (n)-\sqrt{n+b} \log (n+b) \tag 1$$ Assuming that the solution could be large, I used to most basic equivalents $$\sqrt{n+k}\sim \sqrt n+\frac k {2 \sqrt n}\qquad \text{and} \qquad \log(n+k)\sim \log(n)+\frac k n$$ Doing it, I arrive to the problem of solving for $n$ $$n (a-b) \log (n)-2bn -b^2=0$$ the solution of which (if it exists) being $$n_{(1)}=\frac{b^2}{(a-b)\, W\left(\frac{b^2 }{a-b}e^{-\frac{2 b}{a-b}}\right)}$$ which, assuming that the argument of Lambert function is small, can be approximated as $$n_{(2)}=\frac{b^2}{a-b}+e^{\frac{2 b}{a-b}}$$ Now, the problem is that, using series $$g(n)=\frac{(a-b) \log (n)-2 b}{2 \sqrt{n}}+O\left(\frac{1}{n^{3/2}}\right)$$ giving $$n_{(3)}=e^{\frac{2 b}{a-b}}$$
The problem, and then my question (and my surprise), is that $n_{(1)}$ and $n_{(2)}$ are much better than $n_{(3)}$. I give in text table some results $$\left( \begin{array}{cccccc} a & b & n_{(1)} & n_{(2)} & n_{(3)} & \text{solution} \\ 2 & 1 & 8.33137 & 8.38906 & 7.38906 & 8.70391 \\ 3 & 2 & 58.4644 & 58.5982 & 54.5982 & 59.3338 \\ 4 & 3 & 412.331 & 412.429 & 403.429 & 413.769 \\ 5 & 4 & 2996.92 & 2996.96 & 2980.96 & 2998.90 \\ 6 & 5 & 22051.5 & 22051.5 & 22026.5 & 22053.9 \\ 7 & 6 & 162791. & 162791. & 162755. & 162794. \\ & & & & & \\ 3 & 1 & 3.18097 & 3.21828 & 2.71828 & 3.47785 \\ 4 & 2 & 9.18632 & 9.38906 & 7.38906 & 9.78682 \\ 5 & 3 & 24.1918 & 24.5855 & 20.0855 & 25.2064 \\ 6 & 4 & 62.1043 & 62.5982 & 54.5982 & 63.6493 \\ 7 & 5 & 160.439 & 160.913 & 148.413 & 162.586 \\ 8 & 6 & 421.049 & 421.429 & 403.429 & 423.813 \\ & & & & & \\ 4 & 1 & 2.25763 & 2.28107 & 1.94773 & 2.52323 \\ 5 & 2 & 4.96291 & 5.12700 & 3.79367 & 5.45027 \\ 6 & 3 & 9.98014 & 10.3891 & 7.38906 & 10.7364 \\ 7 & 4 & 19.0435 & 19.7252 & 14.3919 & 20.1512 \\ 8 & 5 & 35.4581 & 36.3650 & 28.0316 & 37.0121 \\ 9 & 6 & 65.564 & 66.5982 & 54.5982 & 67.6543 \end{array} \right)$$
I wonder if I am making a mistake somewhere. Comments and suggestions will really be welcomed. Thanks in advance.
We obtained the following approximate equations for $n$:
But it turned out that when we take more precise equivalents we obtain a different equation. Namely, we have
$$\sqrt{n+k} \sim n^{1/2}+\frac 12kn^{-1/2}-\frac 18k^2n^{-3/2},$$
$$\log (n+k)\sim\log n+kn^{-1}-\frac 12k^2n^{-2}.$$
Then
$$g(n)\sim\left(n^{1/2}+\frac 12an^{-1/2}-\frac 18a^2n^{-3/2}\right)\log n- \left(n^{1/2}+\frac 12bn^{-1/2}-\frac 18b^2n^{-3/2}\right) \left(\log n+bn^{-1}-\frac 12b^2n^{-2}\right)\sim$$ $$\frac 12(a-b)n^{-1/2}\log n-bn^{-1/2}-\frac 18(a^2-b^2)n^{-3/2}\log n+ O(n^{-5/2}\log n).$$