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
已知abc为正整数,求证a^3+b^3+c^3+1/abc>=2根号3
已知abc为正整数,求证a^3+b^3+c^3+1/abc>=2根号3
数学人气:414 ℃时间:2020-01-28 19:02:49
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