a+b+c=1,证abc(b+c)≤27/1024
a+b+c=1,证abc(b+c)≤27/1024
日期：2022-04-12 11:03:14 人气：1
解:
ab^2c+abc^2
=abc(b+c)
=1/12*(3a)(2b)(2c)(b+c)
=<1/12*[(3a+2b+2c+b+c)/4]^(1/4)
=1/12*[3/4*(a+b+c)]^4
=1/12*(3/4)^4
=27/1024
ab^2c+abc^2
=abc(b+c)
=1/12*(3a)(2b)(2c)(b+c)
=<1/12*[(3a+2b+2c+b+c)/4]^(1/4)
=1/12*[3/4*(a+b+c)]^4
=1/12*(3/4)^4
=27/1024