04/05/12 00:41
>116
G≡（Π[k=1 to n]a_k）^(1/n) とおくとき、
G>1 ⇒ Σ[k=1 to n] 1/(1+a_k) ≧ n/(1+G)
G<1 ⇒ Σ[k=1 to n] 1/(1+a_k) ≦ n/(1+G)
Equality: a_k=const or G=1.
04/05/12 00:41
>116
G≡（Π[k=1 to n]a_k）^(1/n) とおくとき、
G>1 ⇒ Σ[k=1 to n] 1/(1+a_k) ≧ n/(1+G)
G<1 ⇒ Σ[k=1 to n] 1/(1+a_k) ≦ n/(1+G)
Equality: a_k=const or G=1.