已知x,y,z>0,且x+y+z=1.求证
(1-x)^2/(1+x)+(1-y)^2/(1-y)+(1-y)^1/(1-z)>=√3(x^2+y^2+z^2)
(1-x)^2/(1+x)=(y+z)^2/(1+x)
可以有以下不等式：
(y+z)^2/(1+x)+9(1+x)(y+z)^2/16≥2√{[(y+z)^2/(1+x)][9(1+x)(y+z)^2/16]}
=3(y+z)^2/2
即(y+z)^2/(1+x)≥3(y+z)^2/2-9(1+x)(y+z)^2/16
=(3/16)(y+z)^2[8-3(1+x)]=(3/16)(y+z)^2(5-3x)
=15(x+y)^2/16-9x(y+z)^2/16
所以有
(y+z)^2/(1+x)≥15(x+y)^2/16-9x(y+z)^2/16
如此可得另外两个同型不等式，三式相加可得...全部
(1-x)^2/(1+x)=(y+z)^2/(1+x)
可以有以下不等式：
(y+z)^2/(1+x)+9(1+x)(y+z)^2/16≥2√{[(y+z)^2/(1+x)][9(1+x)(y+z)^2/16]}
=3(y+z)^2/2
即(y+z)^2/(1+x)≥3(y+z)^2/2-9(1+x)(y+z)^2/16
=(3/16)(y+z)^2[8-3(1+x)]=(3/16)(y+z)^2(5-3x)
=15(x+y)^2/16-9x(y+z)^2/16
所以有
(y+z)^2/(1+x)≥15(x+y)^2/16-9x(y+z)^2/16
如此可得另外两个同型不等式，三式相加可得
(y+z)^2/(1+x)+(z+x)^2/(1+y)+(x+y)^2/(1+z)
≥(15/16)[(y+z)^2+(z+x)^2+(x+y)^2]
-(9/16)[x(y+z)^2+y(z+x)^2+z(x+y)^2] (*)
而x(y+z)^2+y(z+x)^2+z(x+y)^2=x+y+z-2(x^2+y^2+z^2)+x^3+y^3+z^3
(y+z)^2+(z+x)^2+(x+y)^2=x^2+y^2+z^2+1
(*)化为(y+z)^2/(1+x)+(z+x)^2/(1+y)+(x+y)^2/(1+z)
≥(15/16)(x^2+y^2+z^2+1)
-(9/16)(x+y+z-2(x^2+y^2+z^2)+x^3+y^3+z^3)
=(33/16)(x^2+y^2+z^2)+15/16-(9/16)(x+y+z+x^3+y^3+z^3)
=(28/16)(x^2+y^2+z^2)+(5/16)(x^2+y^2+z^2)+6/16
-(9/16)(x^3+y^3+z^3)
>(28/16)(x^2+y^2+z^2)+(9/16)[(x^2+y^2+z^2)-(x^3+y^3+z^3)]
>(28/16)(x^2+y^2+z^2)>√3(x^2+y^2+z^2)
由此看出，√3不是最佳上限！
我只好重新证明了！
(1-x)^2/(1+x)=[(1+x)^2-4x]/(1+x)
=1+x-4[1-1/(1+x)]=x-3+4/(1+x)
所以原式左边=x+y+z-9+4[1/(1+x)+1/(1+y)+1/(1+z)]
=-8+4[1/(1+x)+1/(1+y)+1/(1+z)]
原不等式化为
4[1/(1+x)+1/(1+y)+1/(1+z)]-8≥√[3(x^2+y^2+z^2)]
也可以表示为
4-4[x/(1+x)+y/(1+y)+z/(1+z)]≥√[3(x^2+y^2+z^2)]。
收起
