if $n>20202$,show that $$\sum_{j_{1}=1}^{n}\sum_{j_{2}=1}^{n}\cdots\sum_{j_{2020}=1}^{n}\gcd(j_{1},j_{2},\cdots,j_{2020})\le\dfrac{\zeta(2019)}{\zeta(2020)}n^{2020}$$ where $\zeta(s)=\sum_{n=1}^{+\infty}\dfrac{1}{n^s},s>1$
I try:maybe use $\gcd(x,y)=\dfrac{xy}{[x,y]}$,and
$$\forall s>1,\qquad \sum_{m,n\geq 1}\frac{1}{\text{lcm}(m,n)^s} = \sum_{m\geq 1}\frac{g(m)}{m^s}=\color{red}{\frac{\zeta(s)^3}{\zeta(2s)}}.$$ through Euler's product.