Let $gcd(a,b)=(a,b), lcm(a,b)=[a,b]$. if $n=2$, $$ab=(a,b)[a,b]$$ $n=3$, $$\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}=\frac{[a,b,c]^2}{[a,b][b,c][c,a]}$$ I know its proof. Let $$a=\prod_{i=1}^{s}p_i^{\alpha_i},b=\prod_{i=1}^{s}p_i^{\beta_i},c=\prod_{i=1}^{s}p_i^{\gamma_i}$$ For $1\leq i\leq s$, $$2max(\alpha_i,\beta_i,\gamma_i)+\sum_{cyc}{min(\alpha_i,\beta_i)}=2min(\alpha_i,\beta_i,\gamma_i)+\sum_{cyc}{max(\alpha_i,\beta_i)}$$Since this equation is symmetric, we may assume that $\alpha_i\geq\beta_i\geq\gamma_i$ and prove easily.
But I can't extend that for $n$ integers. It has strange logic. I found if $n=4$, $$\sum_{sym}max(\alpha_i,\beta_i,\gamma_i)+\sum_{sym}min(\alpha_i,\beta_i)=\sum_{sym}min(\alpha_i,\beta_i,\gamma_i)+\sum_{sym}max(\alpha_i,\beta_i)$$ $n=5$, $$2\sum_{sym}max(\alpha_i,\beta_i,\gamma_i)+3\sum_{sym}min(\alpha_i,\beta_i)=2\sum_{sym}min(\alpha_i,\beta_i,\gamma_i)+3\sum_{sym}max(\alpha_i,\beta_i)$$ But $n=6$, $$\sum_{sym}max(\alpha_i,\beta_i,\gamma_i)+2\sum_{sym}min(\alpha_i,\beta_i)\neq\sum_{sym}min(\alpha_i,\beta_i,\gamma_i)+2\sum_{sym}max(\alpha_i,\beta_i)$$ Can anyone help me?
