Có: 

`xyz+xy+yz+zx=(1+x)(x^2+xy+yz+zx)-x^2(1+x+y+z)=(1+x)(x+z)(x+y)-x^2(1+x+y+z)`

`=> x+y+z+1=(1+x)(x+z)(x+y)-x^2(1+x+y+z)`

`=>(x+y+z+1)(1+x^2)=(1+x)(x+z)(x+y)`

`=>(1+x^2)/(1+x)=((x+z)(x+y))/(x+y+z+1)`

CMTT `=> VT=sum sqrt{((x+z)(x+y))/(x+y+z+1)}`

Áp dụng Bunhia:  `(sum sqrt{((x+z)(x+y))/(x+y+z+1)})^2<=(sum (x+y)/(1+x+y+z))(sum (x+z))`

`=> VT^2<=((2 sum x)/(1+sum x))(2 sum x)=4 (sum x)^2/(1+sum x)`

Đặt `sum x =a`

`=> VT^2<= 4a^2/(1+a)`

Cần cm: `4a^2/(1+a) <= 9(a/3)^(5/4)`

`<=>(4a^2/(1+a))^4<= 9^4(a/3)^5`

`<=>4^4.a^8<=27a^5(1+a)^4`

`<=>256a^3<=27(1+a)^4`

`<=> 27a^4-148a^3+162a^2+108a+27>=0`

`<=> (27a^2 + 14a + 3)(a - 3)^2>=0` (luôn đúng)

`=> đpcm`