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

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

• Just use $n=\sum_{d|n} \phi(d)$ and exchange sums. The inequality drops at once. – Lucia Apr 22 '20 at 3:45
• Thanks, exchange the sums maybe is not easy to prove?Thanks – function sug Apr 22 '20 at 3:48
• No, it is easy to prove. Try, and throw out all that you wrote about the case k=2. And this doesn't seem like a research level problem -- where does it come from? – Lucia Apr 22 '20 at 3:55
• oh,I don't understand，can you post your answer,Thanks,this problem is from web – function sug Apr 22 '20 at 4:23

Following Lucia's hint, we have that $$S_k(n)=\frac{1}{n^k}\sum_{d=1}^\infty\phi(d)\left\lfloor\frac{n}{d}\right\rfloor^k \leq\sum_{d=1}^\infty\frac{\phi(d)}{d^k} =\frac{\zeta(k-1)}{\zeta(k)}.$$
• Thanks,the last $=$ why? – function sug Apr 23 '20 at 6:42