(2)原式即証\(\displaystyle (1+r_1^3)(1+r_2^3)(1+r_3^3)\geq (1+r_1r_2r_3)^3\)
其中\(\displaystyle r_1=\frac{a_2}{a_1},r_2=\frac{b_2}{b_1},r_3=\frac{c_2}{c_1}\)
\(\displaystyle (1+r_1^3)(1+r_2^3)(1+r_3^3)=1+(r_1^3+r_2^3+r_3^3)+[(r_1r_2)^3+(r_2r_3)^3+(r_3r_1)^3]+(r_1r_2r_3)^3\geq 1+3r_1r_2r_3+3(r_1r_2r_3)^2+(r_1r_2r_3)^3=(1+r_1r_2r_3)^3\)得証
[ 本帖最後由 satsuki931000 於 2021-5-7 21:20 編輯 ]