一道高中数学题，帮忙解一下 证:xyz(x+y)(x+z)(y+z)≤(x눀+y눀)(x눀+z눀)

这里应该先证明一下：k≠0，则有k+1/k<=k^2+1/k^2；
k^2+1/k^2-(k+1/k)=k^2-k-(1/k-1/k^2)=k(k-1)-(1-1/k)/k=k(k-1)-(k-1)/k^2=(k-1)(k-1/k^2)
=(k-1)(k^3-1)/k^2=(k-1)^2(k^2+k+1)/k^2=(k-1)^2[(k+1/2)^2+3/4]/k^2>=0

xyz(x+y)(x+z)(y+z)=xyz[x^2+(y+z)x+yz](y+z)=xyz[x^2(y+z)+(y+z)^2x+yz(y+z)]
=xyz(x^2y+x^2z+y^2x+2xyz+z^2x+y^2z+z^2y)
=xyz[(x^2y+xy^2)+(z^2x+x^2z)+(y^2z+z^2y)+2xyz].................(1)

(x^2+y^2)(x^2+z^2)(y^2+z^2)=(x^4y^2+y^4x^2)+(x^4z^2+z^4x^2)+(y^4z^2+z^4y^2)+2x^2y^2z^2
................(2)

如果x,y,z其中至少有一个为0,比如x=0,则(2)=y^2z^2(y^2+z^2)>=0=(1),不等式成立,
所以设xyz≠0，则x^2y^2z^2>0
[(1)-(2)]/(x^2y^2z^2)
=[(x/z+y/z)+(z/y+x/y)+(y/x+z/x)]-[(x/z)^2+(y/z)^2+(z/y)^2+(x/y)^2+(y/x)^2+(z/x)^2]
=(x/z+z/x)-[(x/z)^2+(z/x)^2]+(x/y+y/x)-[(x/y)^2+(y/x)^2]+(y/z+z/y)-[(y/z)^2+z/y)^2]
<=0,所以(1)<=(2),不等式成立,

∵[(x+y)/2]²≤(x²+y²)/2
[(y+z)/2]²≤(y²+z²)/2
[(z+x)/2]²≤(z²+x²)/2
∴化简得[(x+y)(y+z)(z+x)]²/8≤(x²+y²)(y²+z²)(z²+x²)
对比题中不等式,只要证明(x+y)(y+z)(z+x)≥8xyz即可
∵x+y≥2√(xy),y+z≥2√(yz),z+x≥2√(zx)
∴(x+y)(y+z)(z+x)≥8√(xy*yz*zx)=8xyz
上述所有不等式的等号都在x=y=z处取得
∴原不等式成立