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][1]) 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]

  [1]: http://enaslund.wordpress.com/2012/01/15/the-sum-of-the-totient-function-and-montgomerys-lower-bound/
[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