if $n>k>1$ be postive integer,show that


$$S\_{k}(n)=\dfrac{1}{n^k}\sum\_{j\_{1}=1}^{n}\sum\_{j\_{2}=1}^{n}\cdots\sum\_{j\_{k}=1}^{n}\gcd(j\_{1},j\_{2},\cdots,j\_{k})\le\dfrac{\zeta(k-1)}{\zeta(k)} \tag{1}$$


where $\zeta(s)=\sum\_{n=1}^{+\infty}\dfrac{1}{n^s},s>1$


I have known this $S\_{2}(n)$some approximation reslut,such as following
$$
\begin{align}
S\_{2}(n)&=\frac{1}{n^2}\sum\_{i=1}^n\sum\_{j=1}^n\gcd(i,j) \\ &= \frac{1}{n^2}\sum\_{g=1}^n\sum\_{i\le\lfloor n/g\rfloor}\sum\_{\substack{j\le\lfloor n/g\rfloor\\(i,j)=1}} g \\
&= \frac{1}{n^2}\sum\_{g=1}^n g\left(-1+2\sum\_{i=1}^{\lfloor n/g\rfloor} \varphi(i)\right) \\
&= -\frac{n(n+1)}{2n^2}+\frac{2}{n}\sum\_{g=1}^n \frac{g}{n}\sum\_{i=1}^{\lfloor n/g\rfloor} \varphi(i)
\end{align}$$
Write
$$
f(x) = \frac{1}{x}\sum\_{i\le x}\varphi(i) = \frac{3x}{\pi^2}+E(x) \\
E(x) = o(\log x)
$$
(see [Eric Naslund's exposition](http://enaslund.wordpress.com/2012/01/15/the-sum-of-the-totient-function-and-montgomerys-lower-bound/)) then
$$
\begin{align}
S\_{2}(n) &= -\frac{1}{2}-\frac{1}{2n}+\frac{2}{n}\sum\_{g=1}^{n}f(n/g) \\
&= -\frac{1}{2}-\frac{1}{2n}+\frac{6}{\pi^2}\sum\_{g=1}^{n}\frac{1}{g}+\frac{2}{n}\sum\_{g=1}^n E(n/g) \\
&= \frac{6}{\pi^2}\log n+\frac{6\gamma}{\pi^2}-\frac{1}{2}+C+o(1) \\
&= \frac{6}{\pi^2}\log n + C' + o(1)
\end{align}
$$
where the constant $C$ arises from
$$
E(x) = o(\log x) \\
\left|\frac{2}{n}\sum\_{g=1}^n E(n/g)\right|< \frac{C}{n}\sum\_{g=1}^n\log(n/g)=C\left(\log n - \frac{\log n!}{n}\right)=C+o(1)
$$
by Stirling's approximation. Calculations suggest $C=0.39344\cdots, C'=0.24434\cdots$.
also see:[2](https://mathoverflow.net/questions/314998/on-approximation-of-sum-a-b-1n-gcda-b?rq=1)


But for $(1)$inequality, there exist some reslut?or anyone can help prove,Thanks



inequality for sum $\sum_{j_{i}=1,i=1,\cdots,k}^{n}\gcd(j_{1},j_{2},\cdots,j_{k})$
Following Lucia's hint, we have that
$$S\_k(n)=\frac{1}{n^k}\sum\_{d=1}^\infty\phi(d)\left\lfloor\frac{n}{d}\right\rfloor^k
\leq\sum\_{d=1}^\infty\frac{\phi(d)}{d^k}
=\frac{\zeta(k-1)}{\zeta(k)}.$$


