inequality for sum $\sum_{j_{i}=1,i=1,\cdots,k}^{n}\gcd(j_{1},j_{2},\cdots,j_{k})$

Question

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) 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

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

Answer 1

Following Lucia's hint, we have that $$S_k(n)=\frac{1}{n^k}\sum_{d=1}^\infty\phi(d)\left\lfloor\frac{n}{d}\right\rfloor^k \leq\sum_{d=1}^\infty\frac{\phi(d)}{d^k} =\frac{\zeta(k-1)}{\zeta(k)}.$$