题目

设a,b,c满足ab+bc+cd+da=1,求证：a^3/(b+c+d)+b^3/(a+c+d)+c^3/(a+b+d)+d^3/(a+b+c)大于等于1/3

提问时间：2020-11-13

答案

若a=2 b=-1/2 c=-4 d=0
满足ab+bc+cd+da=1
a^3/(b+c+d)+b^3/(a+c+d)+c^3/(a+b+d)+d^3/(a+b+c)=0
由平均值不等式
a^3/(b+c+d)+[a(b+c+d)]/9>=2a^2/3
同理b^3/(a+c+d)+[b(a+c+d)]/9>=2b^2/3
c^3/(a+b+d)+[c(a+b+d)]/9>=2c^2/3
d^3/(a+b+c)+[d(a+b+c)]/9>=2d^2/3
a^3/(b+c+d)+b^3/(a+c+d)+c^3/(a+b+d)+d^3/(a+b+c)
>=(2/3)(a^2+2b^2+2c^2+2d^2)-[a(b+c+d)+b(a+c+d)+c(a+b+d)+d(a+b+c)]/9
=(2/3)(a^2+2b^2+2c^2+2d^2)-[2+2(ac+bd)]/9
>=(2/3)(a^2+2b^2+2c^2+2d^2)-(2+a^2+b^2+c^2+d^2)/9
=(5/9)(a^2+b^2+c^2+d^2)-2/9
>=(5/9)(ab+bc+cd+da)-2/9
=1/3
取等号时a=b=c=d=1/2