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设a,b,c,d是正实数,证明:a+b+c+d/abcd≤1/a^3+1/b^3+1/c^3+1/d^3

来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/18 11:44:02
设a,b,c,d是正实数,证明:a+b+c+d/abcd≤1/a^3+1/b^3+1/c^3+1/d^3
(1/(3a^3)+1/(3b^3)+1/(3c^3))/3>=三次根号(1/(3a^3)*1/(3b^3)*1/(3c^3))=1/(3abc)
1/(3a^3)+1/(3b^3)+1/(3c^3)>=1/(abc)=d/abcd
同理
1/(3a^3)+1/(3c^3)+1/(3d^3)>=1/(acd)=b/abcd
1/(3a^3)+1/(3b^3)+1/(3c^3)>=1/(abd)=c/abcd
1/(3b^3)+1/(3c^3)+1/(3d^3)>=1/(bcd)=a/abcd
四式相加,得证