(ab+a+b+1)(ab+ac+bc+c^2)=(a+1)(b+1)(a+c)(b+c).
(a+1)×(b+1)×(a+c)×(b+c)≥2√a×2√b×2√ac×2√bc=16abc,
(ab+a+b+1)(ab+ac+bc+c^2)≥16abc.
(ab+a+b+1)(ab+ac+bc+c平方)>=16abc,已知abc都为正整数,
(ab+a+b+1)(ab+ac+bc+c平方)>=16abc,已知abc都为正整数,
是要证明那个>=16abc成立
是要证明那个>=16abc成立
数学人气:664 ℃时间:2020-01-29 13:35:53
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