原式 = (3a³+3b³+3c³+3/abc)/3
= (3a³+3b³+3c³+1/abc+1/abc+1/abc)/3
≥ 6*(3a³*3b³*3c³*1/abc*1/abc*1/abc)^(1/6) / 3
= 6*√3 / 3
= 2√3