Cho ba số thực dương $x,y,z$ thỏa mãn $xyz=1$. Tìm giá trị nhỏ nhất của biểu thức $P=\frac{x+2}{{{x}^{3}}(y+z)}+\frac{y+2}{{{y}^{3}}(z+x)}+\frac{z+2}{{{z}^{3}}(x+y)}.$

`P=(x+2)/(x^3(y+z))+(y+2)/(y^3(z+x))+(z+2)/(z^3(x+y))`

`P=x/(x^3(y+z))+y/(y^3(z+x))+z/(z^3(x+y))+2/(x^3(y+z))+2/(y^3(z+x))+2/(z^3(x+y))`

`P=1/(x^2(y+z))+1/(y^2(z+x))+1/(z^2(x+y))+2[1/(x^3(y+z))+1/(y^3(z+x))+1/(z^3(x+y))]`

`P=(xyz)/(x^2(y+z))+(xyz)/(y^2(z+x))+(xyz)/(z^2(x+y))+2[(xyz)^2/(x^3(y+z))+(xyz)^2/(y^3(z+x))+(xyz)^2/(z^3(x+y))]`

`P=(yz)/(x(y+z))+(zx)/(y(z+x))+(xy)/(z(x+y))+2[(yz)^2/(x(y+z))+(zx)^2/(y(z+x))+(xy)^2/(z(x+y))]`

`P=(yz)/(xy+zx)+(zx)/(yz+xy)+(xy)/(zx+yz)+2[(yz)^2/(xy+zx)+(zx)^2/(yz+xy)+(xy)^2/(zx+yz)]`

Đặt `xy=a, yz=b, zx=c` ta được:

`P=b/(a+c)+c/(b+a)+a/(c+b)+2(b^2/(a+c)+c^2/(b+a)+a^2/(c+b))`

Ta có:

`b/(a+c)+c/(b+a)+a/(c+b)>=3/2`

`b^2/(a+c)+c^2/(b+a)+a^2/(c+b)>=(a+b+c)/2>=(3\root{3}{abc})/2=(3\root{3}{1})/2=3/2`

`=>P>=3/2+2. 3/2=9/2`

Dấu $"="$ xảy ra

`<=>{(x=y=z),(xyz=1):}`

`<=>x=y=z=1`

Vậy `minP=9/2` khi `x=y=z=1`