a,b,c都是正数,ab+bc+ca=1则a+b+c<=1/(3abc)为何错了

ab + bc + ca = 1
(ab + bc + ca)^2
= (ab + bc + ca)(ab + bc + ca)
= a^2b^2 + 2acb^2 + 2a^2bc + b^2c^2 + 2abc^2 + a^2c^2
= 1

假设 a+b+c<=1/(3abc) 对了
a^2bc + b^2ac + c^2ab <= 1/3
(1 - a^2b^2 - b^2c^2 - c^2a^2)/2 <=1/3
(1 - a^2b^2 - b^2c^2 - c^2a^2) <= 2/3
a^2b^2 + b^2c^2 + c^2a^2 >= 1/3

a^2b^2 + b^2c^2 + c^2a^2
= 1/2(a^2b^2 + b^2c^2 + c^2a^2 + a^2b^2 + b^2c^2 + c^2a^2 )
>= acb^2 + bca^2 + abc^2
当 a^2b^2 = b^2c^2 = c^2a^2 时等号成立
ab=bc=ca, a=c=b
ab+bc+ca=1, 3a^2=1, a^2=1/3
acb^2 + bca^2 + abc^2 = 1/9 * 3 = 1/3

所以没错！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！1