Cho các số thực $x,y,z$ thoả mãn điều kiện $x\geq 1;y\geq 2;z\geq3 $ và 
$\frac{x^{2}-x+1}{x+\sqrt{x-1}}+\frac{y^{2}-y+2}{y+\sqrt{y-2}}+\frac{z^{2}-z+3}{z+\sqrt{z-3}}=12$
Tìm GTLN,GTNN của $A=x+y+z$
$\frac{x^2-(x-1)}{x+\sqrt{x-1}}+\frac{y^2-(y-2)}{y+\sqrt{y-2}}+\frac{z^2-(z-3)}{z+\sqrt{z-3)}}=12 \\ \Leftrightarrow x-\sqrt{x-1}+y-\sqrt{y-2}+z-\sqrt{z-3}=12 \\ \Leftrightarrow x+y+z-12 =\sqrt{x-1}+\sqrt{y-2}+\sqrt{z-3} (\star)$
Đặt $t=x+y+z (t \ge 12)$
Từ $(\star)\Rightarrow t-12 \le \sqrt{3(t-6)}\Leftrightarrow 6 \le t \le 18\Rightarrow \max_{t \ge 12} t=18$
Từ $(\star)\Rightarrow (t-12)^2=t-6+2\left(\sqrt{(x-1)(y-2)}+\sqrt{(y-2)(z-3)}+\sqrt{(z-3)(x-1} \right)$
$\ge t-6\Leftrightarrow t^2-24t+144 \ge t-6\Leftrightarrow\left[ \begin{array}{l} t \ge 15\\ t \le 10 \end{array} \right.\Rightarrow \min_{t\ge 12} t=15$
$KL: GTLN=18\Leftrightarrow (x,y,z)=(5,6,7)$
$GTNN=15\Leftrightarrow (x,y,z)=\{(1;2;12);(1,11,3);(10,2,3)\}$



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