How prove this inequality $\frac{3}{4}\le \left(\frac{1}{n}\right)^{\frac{1}{n-1}}+\left(\frac{1}{n}\right)^{\frac{n}{n-1}}<1$ 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 
 A: 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 $n\geqslant2$,
$$
\frac34=\mathrm e^{u(2)}\leqslant\left(\dfrac{1}{n}\right)^{\frac{1}{n-1}}+\left(\dfrac{1}{n}\right)^{\frac{n}{n-1}}\lt\mathrm e^{u(+\infty)}=1.
$$
A: 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.
