令 a = x + y , b = y + z , c = z + x.(由于是三角形三边长,肯定能找到相应的正实数x,y,z满足条件.)因为a+b+c=1所以 x + y + z = 1/2. 注意到此时有平均值不等式 xyz <= 1/216因此 F = a^2 + b^2 + c^2 + 4abc = (x+y)^2 + (y+z)^2 + (z+x)^2 + 4(x+y)(y+z)(z+x) = 2(x^2 + y^2 + z^2) + 2(xy + yz + zx) + 4(1/2 - x)(1/2 - y)(1/2 - z) = 2(x^2 + y^2 + z^2) + 2(xy + yz + zx) + 4(1/8 - 1/4(x+y+z) + 1/2(xy + yz + zx) - xyz)= 2(x^2 + y^2 + z^2) + 2(xy + yz + zx) + 2(xy + yz + zx) - 4xyz= 2(x+y+z)^2 - 4xyz= 1/2 - 4xyz因为Xyz都大于0至此, F < 1/2 得证