x2−(x−1)x+√x−1+y2−(y−2)y+√y−2+z2−(z−3)z+√z−3)=12⇔x−√x−1+y−√y−2+z−√z−3=12⇔x+y+z−12=√x−1+√y−2+√z−3(⋆)Đặt t=x+y+z(t≥12)
Từ (⋆)⇒t−12≤√3(t−6)⇔6≤t≤18⇒max
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)\}