Prove that ${\sqrt {n} }^{\sqrt {n+1}} > {\sqrt {n+1}}^{\sqrt {n}};n=7,8,9...$

Question

Prove that ${\sqrt {n} }^{\sqrt {n+1}} > {\sqrt {n+1}}^{\sqrt {n}};n=7,8,9...$

SOURCE : "Inequalities proposed in CRUX Mathematicorum"

I tried induction, but could not prove it.

Can anybody provide a hint ?

Answer 1

One can check the case $n=7$ by hand.

Taking logarithms on both sides, the inequality becomes : $$\sqrt{n+1}\log \sqrt n > \sqrt{n}\log \sqrt{n+1} $$ $${\sqrt{n+1}\over \log \sqrt{n+1}} > {\sqrt n \over \log \sqrt n} $$ So, it is enough to prove $f(x) = \frac{x}{\log x}$ is increasing for $x\ge \sqrt 8$, the easiest way to do that is showing $f'(x)>0$ for $x\ge \sqrt 8$.

Using the quotient rule, $$f'(x)= {\log x-1\over (\log x)^2}$$ , which is obviously positive as $\sqrt 8> e$.

Answer 2

Let $f(x)=2\ln(\ln x)-\ln x$ where $x\in(e^2,\infty)$, then $$f'(x)=\dfrac{1}{x}\Big(\frac{2}{\ln x}-1\Big)<0$$ so \begin{eqnarray} 2\ln(\ln(n+1))-\ln (n+1)&<&2\ln(\ln n)-\ln n\\ 2\ln(\ln(n+1))-2\ln(\ln n)&<&\ln (n+1)-\ln n\\ 2\ln\frac{\ln(n+1)}{\ln n}&<&\ln\frac{n+1}{n}\\ \frac{\ln(n+1)}{\ln n}&<&\frac{\sqrt{n+1}}{\sqrt{n}}\\ \sqrt{n}\ln(n+1)&<&\sqrt{n+1}\ln n\\ (n+1)^\sqrt{n}&<&n^\sqrt{n+1}\\ \sqrt{n+1}^\sqrt{n}&<&\sqrt{n}^\sqrt{n+1} \end{eqnarray}