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小华和小明正在做一道“应用不等式求最值”的习题:已知a,b,c均为正数,且a+b+c=1,求ab~2c~3的最大值。小华解:∵a+b+c=a+ b/2+b/2+c/3+c/3+c/3≥6((a(b/2)~2(c/3)~3)~(1/6)) ∴1≥6((ab~2c~3)~(1/6))/108)),即ab~2~3≤1/432. ∴ab~2c~3的最大值为1/432。小明解:根据a+b+b+c+c+c≥6((ab~2c~3)~(1/6)),当且仅当a=b=c时取等号,右式最大。又∵a+b+c=1,∴a=b=c=1/3。得ab~2c~3=1/729,既ab~2c~3的最大值为两1/729。小华看着小明的结果,诧异地说:“我们都为都是应用正数的算术平均≥几何平均’,结果怎么不同呢?”小
Xiaohua and Xiaoming are doing exercises to “apply the inequality for the best value”: Know that a, b, c are all positive, and a+b+c=1, find the maximum of ab~2c~3. Xiaohua Solution: ∵a+b+c=a+b/2+b/2+c/3+c/3+c/3≥6((a(b/2)~2(c/3)~3 )~(1/6)) ∴1≥6((ab~2c~3)~(1/6))/108)), ie ab~2~3≤1/432. ∴ab~2c~3 The maximum is 1/432. Xiao Ming solution: According to a+b+b+c+c+c≥6((ab~2c~3)~(1/6)), if and only if a=b=c, the equal sign will be used. . Again ∵a+b+c=1, ∴a=b=c=1/3. To get ab~2c~3=1/729, the maximum value of ab~2c~3 is two 1/729. Xiao Hua looked at Xiaoming’s results and said bizarrely: “We all apply positive numbers with arithmetic mean ≥ geometric mean. How are the results different?”