For any postive integer number $n\ge 2$, show that


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 $n\geqslant2$, $$ \frac34=\mathrm e^{u(2)}\leqslant\left(\dfrac{1}{n}\right)^{\frac{1}{n-1}}+\left(\dfrac{1}{n}\right)^{\frac{n}{n-1}}\lt\mathrm e^{u(+\infty)}=1. $$

  • $\begingroup$ Thank you,It's nice.and for this right,use Bernoulli inequality:w we have $$\le 1+\dfrac{1}{n-1}(\dfrac{1}{n}-1)+\dfrac{n}{n-1}(\dfrac{1}{n}-1)=1$$ $\endgroup$ – math110 Oct 20 '13 at 17:24

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.

  • $\begingroup$ This proves the inequality for $n$ large enough, not for every $n\geqslant2$. $\endgroup$ – Did Oct 20 '13 at 19:04
  • $\begingroup$ That's true, but one can check the other cases ($n<6$) by hand. $\endgroup$ – Jack D'Aurizio Oct 21 '13 at 7:41
  • $\begingroup$ Then one should add an actual proof that "the other cases" are only when $n\leqslant5$. $\endgroup$ – Did Oct 21 '13 at 8:03

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