不等式的証明

using AM>=GM
(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)/6>=[(a^2b)(a^2c)(b^2a)(b^2c)(c^2a)(c^2b)]^1/6
(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)/6>=(a^6b^6c^6)^1/6
(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)/6>=abc
(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)>=6abc


2007-02-03 23:02:23 補充：
http://home.netvigator.com/~leeleung/theorem_0004.htmlhere

這是 AM >= GM 的應用.

0.5 ( a + b ) >= (ab)^0.5
0.5 ( b + c ) >= (bc)^0.5
0.5 ( a + c ) >= (ac)^0.5

therefore 0.125 ( a + b ) ( b + c ) ( b + c ) > = (ab*bc*ca)^ 0.5

( a + b ) ( b + c ) ( b + c ) > = 8 abc
and ( a + b ) ( b + c ) ( b + c ) = a^2b+a^2c+b^2a+b^2c+c^2a+c^2b + 2abc
so a^2b+a^2c+b^2a+b^2c+c^2a+c^2b>6abc