令 w=(4a-b-c-d)/a x=b/a y=c/a z=d/a
带入原式得其等价命题
aaa(4a-(b+c+d))^2+aabbb+acccc+ddddd>=aaaa(4a-(b+c+d))+aaabb+aaccc+adddd
拆掉平方 即
12aaaaa+aaa(b+c+d)^2+aabbb+acccc+ddddd>=7aaaaa(b+c+d)+aaabb+aaccc+adddd (1)
排序 aaaaa+aabbb>=aaabb+aaaab
aaaaa+acccc>=aaccc+aaaac
aaaaa+ddddd>=adddd+aaaad
3式相加得
3aaaaa+aabbb+accccc+ddddd>=aaaa(b+c+d)+aaabb+aaccc+adddd (2)
而9aaaaa+aaa(b+c+d)^2>=6aaaa(b+c+d) (3)
(2)(3)式相加得(1)式
所以原命题得证
排序不是很熟练 但绝对是最后一次修改