Cho `3` số dương `x, y,z` thõa mãn `x+y+z=1`. CMR: `\sqrt{\frac{xy}{xy+z}}+\sqrt{\frac{yz}{yz+x}}+\sqrt{\frac{xz}{xz+y}}\leq \frac{3}{2}`

Lời giải:

 Ta có `:` `xy+z=xy+z(x+y+z)=(x+z)(y+z)`

Do đó `:`

`\sqrt[(xy)/(xy+z)]=\sqrt[{xy}/{(x+z)(y+z)}]<=1/2(x/(x+z)+y/(y+z))“text( (1))`

Tương tự ta cũng có `:`

`\sqrt[(zy)/(zy+x)]=\sqrt[{zy}/{(x+y)(x+z)}]<=1/2(y/(x+y)+z/(x+z))“text( (2))`

`\sqrt[(zx)/(zx+y)]=\sqrt[{zx}/{(z+y)(x+y)}]<=1/2(z/(z+y)+x/(x+y))“text( (3))`

Cộng `(1),(2),(3)` theo từng vế ta có `:`

`\sqrt[(xy)/(xy+z)]+\sqrt[(zy)/(zy+x)]+\sqrt[(zx)/(zx+y)]<=3/2`

Vậy `:` `\sqrt[(xy)/(xy+z)]+\sqrt[(zy)/(zy+x)]+\sqrt[(zx)/(zx+y)]<=3/2` luôn đúng.