For any postive integer number $n\ge 2$, show that
$$\dfrac{3}{4}\le\left(\dfrac{1}{n}\right)^{\frac{1}{n-1}}+\left(\dfrac{1}{n}\right)^{\frac{n}{n-1}}<1$$
My try:let $\dfrac{1}{n}=x$, then $$\left(\dfrac{1}{n}\right)^{\frac{1}{n-1}}+\left(\dfrac{1}{n}\right)^{\frac{n}{n-1}}=x^{\frac{x}{1-x}}+x^{\frac{1}{1-x}}(0<x<1)$$ maybe use Young's inequality.
But I can't,and I think this problem have more nice methods,Thank you
Note that, for every $n\geqslant2$, $$ \left(\dfrac{1}{n}\right)^{\frac{1}{n-1}}+\left(\dfrac{1}{n}\right)^{\frac{n}{n-1}}=\frac{n+1}{n^{n/(n-1)}}=\mathrm e^{u(n)}, $$ where, for every $x\gt1$, $$ u(x)=\log(x+1)-\frac{x}{x-1}\log x. $$ Thus, $$ u'(x)=\frac{v(x)}{(x-1)^2},\qquad v(x)=\log(x)-2\frac{x-1}{x+1}. $$ which yields $$ v'(x)=\frac1x-\frac4{(x+1)^2}=\frac{(x-1)^2}{x(x+1)^2}. $$ Now, $v(1)=0$ and $v$ is increasing hence $v\gt0$ and $u$ is increasing on $(1,+\infty)$, in particular, for every $n\geqslant2$, $$ u(2)\leqslant u(n)\lt u(+\infty). $$ Note that $u(2)=\log3-2\log2$ and that $u(x)$ is also $$ u(x)=\log\left(1+\frac1x\right)-\frac{\log x}{x-1}, $$ hence $u(+\infty)=0$. Finally, for every $n\geqslant2$, $$ \frac34=\mathrm e^{u(2)}\leqslant\left(\dfrac{1}{n}\right)^{\frac{1}{n-1}}+\left(\dfrac{1}{n}\right)^{\frac{n}{n-1}}\lt\mathrm e^{u(+\infty)}=1. $$
I think there is a much simpler way. Notice that $\left(\frac{1}{n}\right)^{\frac{1}{n-1}}+\left(\frac{1}{n}\right)^{\frac{n}{n-1}}$ can be expressed as: $$ a_n=\frac{n+1}{n}\left(\frac{1}{n}\right)^{\frac{1}{n-1}},$$ and we only need to prove that the sequence $\{a_n\}_{n\in\mathbb{N}}$ is increasing, since its limit is clearly $1$. $a_{n+1}\geq a_{n}$ is equivalent to: $$ n^{n^2} (n+2)^{n(n-1)} > (n+1)^{2n^2-n-1}, $$ or to: $$\frac{(n+1)^{(n+1)}}{(n+2)^n}>\left(\frac{(n+1)^2}{n(n+2)}\right)^{n^2},$$ Notice that in virtue of the Bernoulli inequality the LHS is greater that $\frac{n+2}{e}$, while the RHS is smaller than $e$, so for any $n\geq 6$ the inequality holds for sure, and we have to check only $n=2,3,4,5$, for which the difference between the LHS and the RHS is indeed positive by hand (or computer) computation.
