左端：
首先注意到[直接計(jì)算]
a^2 + b^2 + 4abc - (1/2)(a+b)^2 - c(a+b)^2 = (1/2) (a-b)^2 (1-2c) >=0
[a、b、c構(gòu)成三角形,故c= (1/2)(a+b)^2 + c^2 + c(a+b)^2
而a+b+c=1,故上式右端化為：
(1/2)(1-c)^2 + c^2 + c(1-c)^2 = (1/2) - (1/2)c^2(1-2c) >= 13/27
[最后一步是將c^2(1-2c)看作c*c*(1-2c),用平均值不等式]
右端：
a、b、c構(gòu)成三角形,故
(a-b)^2 < c^2
=> (a+b)^2 - c^2 < 4ab [配方]
=> (a+b+c)(a+b-c) < 4ab [平方差公式]
=> 1-2c < 4ab [a+b+c=1]
=> 2c-1+4ab > 0 [移項(xiàng)]
=> (1/2)(2c-1)(2c-1+4ab) < 0 [同前,(2c-1) < 0]
=> (1/2)(2c-1)(2c-1+4ab) + (1/2) < (1/2) [兩邊同加(1/2)]
=> (1/2)(2c-1)^2 + (1/2) + 2ab(2c-1) < (1/2) [展開(kāi)]
=> (1-c)^2 + c^2 + 4abc - 2ab < (1/2) [配方、展開(kāi)]
=> (a+b)^2 + c^2 + 4abc - 2ab < (1/2) [a+b+c=1]
即得所求