Đáp án:

Giải thích các bước giải:

`ĐKXĐ`
$\begin{cases}x-1 \geq 0\\y-2 \geq 0\\z-3 \geq 0\\\end{cases}$
`->` $\begin{cases}x \geq 1\\y \geq 2\\z \geq 3\\\end{cases}$
theo đề
$\begin{cases}x+y+z \geq 9\\\sqrt{x-1}+\sqrt{y-2}+\sqrt{z-3}+5x+4y+3z=xy+yz+zx+11\\\end{cases}$
`->` $\begin{cases}x-1+y-2+z-3 \geq 3\\\sqrt{x-1}+\sqrt{y-2}+\sqrt{z-3}+5(x-1)+4(y-2)+3(z-3)=(x-1)(y-2)+(y-2)(z-3)+(z-3)(x-1)+y-2+2x+3y+2z-6+3x+z-3\\\end{cases}$
`->` $\begin{cases}x-2+y-2+z-3 \geq 3\\\sqrt{x-1}+\sqrt{y-2}+\sqrt{z-3}+5(x-1)+4(y-2)+3(z-3)=(x-1)(y-2)+(y-2)(z-3)+(z-3)(x-1)+5(x-1)+4(y-2)+3(z-3)\\\end{cases}$
`->` $\begin{cases}x-2+y-2+z-3 \geq 3\\\sqrt{x-1}+\sqrt{y-2}+\sqrt{z-3}=(x-1)(y-2)+(y-2)(z-3)+(z-3)(x-1)\\\end{cases}$
cần `CM:2sqrt{x-1}>=(y-2)(x-1)+(z-3)(x-1)`
`<=>2sqrt{x-1}>=(x-1)(y-2+z-3)`
`<=>2sqrt{x-1}>=(x-1)(z+y-5)`
`<=>2sqrt{x-1}>=(x-1)(z-3+y-2)`
`<=>2sqrt{x-1}>=(x-1)(3-(x-1))`(do `x-1+y-2+z-3 \ge 3)`
`<=>2sqrt{x-1}>=3(x-1)-(x-1)^2`
`<=>(x-1+2sqrt{x-1})(sqrt{x-1}-1)^2>=0` luôn đúng
CMTT ta có
`2sqrt{y-2}>=(y-2)(z-3)+(y-2)(x-1)`
`2sqrt{z-3}>=(y-2)(z-3)+(z-3)(x-1)`
`->2sqrt{x-1}+2sqrt{y-2}+2sqrt{z-3}>=2[(y-2)(z-3)+(z-3)(x-1)+(x-1)(y-2]`
`->sqrt{x-1}+sqrt{y-2}+sqrt{z-3}>=(y-2)(z-3)+(z-3)(x-1)+(x-1)(y-2)`
dấu = xảy ra khi
$\begin{cases}\left[ \begin{array}{l}x=1\\x=2\end{array} \right. \\\left[ \begin{array}{l}y=2\\y=3\end{array} \right. \\\left[ \begin{array}{l}z=3\\z=4\end{array} \right. \\\end{cases}$
`->` \(\left[ \begin{array}{l}\begin{cases}x=1\\y=2\\z=3\\\end{cases}\\\begin{cases}x=2\\y=3\\z=4\\\end{cases}\end{array} \right.\)

hmu hmu em ms đi học về ạ sorry ad nha :<