证明 bc+ca+ab=[(a+b+c)^2-(a^2+b^2+c^2)]/2=1 设f(x)=x(1-x)^2，只需证明:f(a)=f(b)=f(c) 即可. 因为 (x-a)*(x-b)*(x-c)=x^3-(a+b+c)x^2+(bc+ca+ab)x-abc =x^3-2x^2+x-abc 所以 f(x)=(x-a)*(x-b)*(x-c)+abc (1) 令x=a,b,c得f(a)=f(b)=f(c) ， 从而 a(1-a)^2=b(1-b)^2=c(1-c)^2=abc。
证2：c=2-a-b, a^2+b^2+(2-a-b)^2=2, 2a^2+2b^2+2ab-4a-4b+2=0, a^2+b^2+ab-2a-2b+1=0, a^2=(2-b)a-b^2+2b-1(这是技巧，用于把a降次)，于是 a(1-a)^2=a(1-2a+a^2)=a[1-2a+(2-b)a-b^2+2b-1]=a[-ab-b^2+2b]=b[(2-b)a-a^2]=b[b^2-2b+1]=b(1-b)^2, 同理b(1-b)^2=c(1-c)^2。
证1：c=2-a-b, a^2+b^2+(2-a-b)^2=2, 2a^2+2b^2+2ab-4a-4b+2=0, a^2+b^2+ab-2a-2b+1=0, 设a=(x+y)/√2，b=(-x+y)/√2，则 （1/2)x^2+(3/2)y^2-2√2y+1=0, x^2+3y^2-4√2y+2=0, x^2+3[y-(2√2)/3]^2=2/3, 作三角代换，x=[(√6)/3]cosα，y=√2/3*[2+sinα]。
于是a=[1/3][（√3)cosα+2+sinα]=[2/3][1+sin(α+π/3）], a(1-a)^2=[2/3][1+sin(α+π/3）]{1-[2/3][1+sin(α+π/3）]}^2 =[2/3][1+sin(α+π/3）]{1-(4/3)[1+sin(α+π/3)]+(4/9)[1+sin(α+π/3)]^2} =(2/3){[1+sin(α+π/3)]-(4/3)[1+sin(α+π/3)]^2+(4/9)[1+sin(α+π/3)]^3} =(2/3){1+sin(α+π/3)-(4/3)[1+2sin(α+π/3)+sin^2(α+π/3)]+(4/9)[1+3sin(α+π/3)+3sin^2(α+π/3)+sin^3(α+π/3)]} =(2/3)[1/9-(1/3)sin(α+π/3)+(4/9)sin^3(α+π/3)] =(2/27)[1-sin(3α+π)] =(2/27)(1+sin3α), 同理b=(2/3)[1+sin(α-π/3）] b(1-b)^2=(2/27)(1+sin3α), c=2-(a+b)=2-(2/3)[2+sin(α+π/3)+sin(α-π/3)]=2-(2/3)[2+2sinα]=(2/3)[1+sin(α+π)], c(1-c)^2=(2/27)(1+sin3α)。
答：招生广告哟！ 参加培训肯定有好处！详情>>
