`\color{#FF0000}{T}\color{#FF7F24}{i}\color{#FFFF00}{t}\color{#7FFF00}{u}\color{#00F5FF}{u}\color{#FF1493}{u}`

$b)$ Có `(x+y+z)^2`

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

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

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

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có:

`(1)/(x^2+y^2+z)+(1)/(y^2+z^2+x)+(1)/(z^2+x^2+y)>=((1+1+1)^2)/(x^2+y^2+z+y^2+z^2+x+z^2+x^2+y)=(9)/(2(x^2+y^2+z^2)+(x+y+z))`

`->` Ta cần CM `(9)/(2(x^2+y^2+z^2)+(x+y+z))<=1 (1)`

`<=> 9<=2(x^2+y^2+z^2)+(x+y+z)`

`<=> 9<=2.[(x+y+z)^2-2(x+y+z)]+(x+y+z)`

`<=> 9<=2(x+y+z)^2-3(x+y+z)`

Đặt `x+y+z=a (a>0) -> 2a^2-3a-9>=0`

`-> (a-3)(2a+3)>=0`

`-> a>=3` (vì `a>0 -> 2a+3>0`)

`-> x+y+z>=3`

Lại có: `(x-y)^2+(y-z)^2+(z-x)^2>=0`

`-> 2x^2+2y^2+2z^2-2xy-2yz-2zx>=0`

`-> x^2+y^2+z^2>=xy+yz+zx`

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

`-> x+y+z>=3`

`-> (1)` luôn đúng

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