Inequality $n^{\sqrt{n+2}}>(n+1)^{\sqrt{n+1}}$ solution

Question

How to find the solution of inequality $$n^{\sqrt{n+2}}>(n+1)^{\sqrt{n+1}}$$ for integers with basic analysis without derivatives.

I don't have any idea how to start with it.

Answer 1

You are looking for the smallest $n$ such that

$$\frac{n^{\sqrt{n+2}}}{(n+1)^{\sqrt{n+1}}} > 1$$ Take logarithms and search for the zero of $$g(n)=\sqrt{n+2} \log (n)-\sqrt{n+1} \log (n+1)$$ Assume that $n$ is large and use $$\sqrt{n+2}\sim \sqrt n+\frac 1 { \sqrt n}$$ $$\sqrt{n+1}\sim \sqrt n+\frac 1 {2 \sqrt n}$$ $$\log(n+1)=\log(n)+\log\left(1+\frac 1 n\right)\sim \log(n)+\frac 1 n$$ Replace to get $$g(n) \sim -\frac{1}{2n^{3/2}} (2 n-n \log (n)+1)$$ The solution of $2 n-n \log (n)+1=0$ is given in terms of Lambert function; so, the approximation is $$n =\frac{1}{W\left(\frac{1}{e^2}\right)}$$ Since the argument is small, use $W(x)\approx \frac{x}{x+1}$ which makes $\color{red}{n=1+e^2 \approx 8.39}$ while the exact solution of $g(n)=0$ in the real domain would be $8.70$. Then $\color{red}{n=9}$.

Let us check the values of $$f(n)=n^{\sqrt{n+2}}-(n+1)^{\sqrt{n+1}}$$ $$f(8)=8^{\sqrt{10}}-729 \approx -11.50 \qquad f(9)=9^{\sqrt{11}}-10^{\sqrt{10}}\approx 8.68$$