Is there an easy way to decide the sign of $\sqrt x \ln (x + 1) - \sqrt {x + 1} \ln x$?

Question

Is there an easy way to decide the sign of $\sqrt x \ln (x + 1) - \sqrt {x + 1} \ln x$?

The original problem is to decide when ${(n + 1)^{\sqrt n }}$ is greater than ${n^{\sqrt {n + 1} }}$.

It seems that the function is monotonic and has a unique root near $x=6.9$, but I wonder how to decide its sign explicitly since the derivative is a little bit complex.

Any help will be appreciated:)

Answer 1

When I plot it in Alpha it appears to be monotoniclly decreasing. Given your original problem, you are really interested in its sign. It changes from positive to negative at about $6.91$. As a result, for $n \in \Bbb N$ we have $(n+1)^{\sqrt n} \gt n^{\sqrt{n+1}}$ for $n \le 7$. The inequality reverses for $n \gt 7$

Answer 2

Hint

Consider the function $$f(y)=\frac{\log(y)}{\sqrt y}\implies f'(y)=-\frac{\log (y)-2}{2 y^{3/2}}$$ The derivative cancels when $y=e^2$ and it is positive for $0 < y < e^2$.

Now, use

$$\sqrt x \,\log (x + 1) - \sqrt {x + 1}\, \log (x)={\sqrt x\,\sqrt {x+1}}\left(\frac{\log(x+1)}{\sqrt {x + 1} }-\frac{\log(x)}{\sqrt {x} }\right)$$