Proof that $(\sqrt{n})^{\sqrt{n+1}}>(\sqrt{n+1})^{\sqrt{n}}.$ For values $n= 7,8,9....$
The solution is given as follows. Could anyone help to explain this? (Especially the last step "Consequently...")
Also, the question had mentioned (but had not stated) that the proof by calculus is possible. Could anyone also show this?
Consider that $${\sqrt n}^{\sqrt {n+1}} >{\sqrt {n+1}}^{\sqrt n}\implies {\sqrt {n+1}}\log({\sqrt n} )>{\sqrt n}\log({\sqrt {n+1}} )$$ that is to say $$\frac{\log({\sqrt n} ) } {\sqrt n } >\frac{\log({\sqrt {n+1}} )}{\sqrt {n+1}}$$
In the real domain, consider the function $$f(x)=\frac{\log({\sqrt x} ) } {\sqrt x } \implies f'(x)=-\frac{\log (x)-2}{4 x^{3/2}}$$ The first derivative cancels at $x=e^2$ and the second deriative test shows that this is a maximum.
Now $e^2 \approx 7.38$ could help.