`x\sqrt(1-y^2)+y\sqrt(1-z^2)+z\sqrt(1-x^2)=3/2`

`->2(x\sqrt(1-y^2)+y\sqrt(1-z^2)+z\sqrt(1-x^2))=3/2. 2`

`->2x\sqrt(1-y^2)+2y\sqrt(1-z^2)+2z\sqrt(1-x^2)=3`

`->(1-2x\sqrt(1-y^2))+(1-2y\sqrt(1-z^2))+(1-2z\sqrt(1-x^2))=0`

`->(x^2-2x\sqrt(1-y^2)+1-y^2)+(y^2-2y\sqrt(1-z^2)+1-z^2)+(z^2-2z\sqrt(1-x^2)+1-x^2)=0`

`->[x^2-2.x.\sqrt(1-y^2)+(\sqrt(1-y^2))^2]+[y^2-2.y.\sqrt(1-z^2)+(\sqrt(1-z^2))^2]+[z^2-2.z.\sqrt(1-x^2)+(\sqrt(1-x^2))^2]=0`

`->(x-\sqrt(1-y^2))^2+(y-\sqrt(1-z^2))^2+(z-\sqrt(1-x^2))^2=0`

Ta có: `{((x-\sqrt(1-y^2))^2>=0AAx;y),((y-\sqrt(1-z^2))^2>=0AAy;z),((z-\sqrt(1-x^2))^2>=0AAz;x):}`

`->(x-\sqrt(1-y^2))^2+(y-\sqrt(1-z^2))^2+(z-\sqrt(1-x^2))^2>=0`

Dấu "`=`" `<=>{(x-\sqrt(1-y^2)=0),(y-\sqrt(1-z^2)=0),(z-\sqrt(1-x^2)=0):}`

`<=>{(x=\sqrt(1-y^2)),(y=\sqrt(1-z^2)),(z=\sqrt(1-x^2)):}`

`<=>{(x^2=1-y^2(1)),(y^2=1-z^2(2)),(z^2=1-x^2(3)):}`

Cộng vế với vế `(1),(2),(3)` ta được: `x^2+y^2+z^2=3-(x^2+y^2+z^2)`

`->2(x^2+y^2+z^2)=3`

`->x^2+y^2+z^2=3/2`

Vây `...`