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Đặt $x-1=a,y-1=b,z-1=c,a,b,c>0$
Suy ra: $x=a+1,y=b+1,z=c+1$.
Giả thiết trở thành:
$(a+1)(b+1)+(b+1)(c+1)+(c+1)(a+1)\ge2(a+1)(b+1)(c+1)$
$\Leftrightarrow ab+bc+ca+2(a+b+c)+3\ge2(abc+ab+bc+ca+a+b+c+1)$
$\Leftrightarrow 2abc+ab+bc+ca\le1$
Đặt $t=\sqrt[3]{abc}$
Áp dụng BĐT Cauchy ta có:
$1\ge2abc+ab+bc+ca$
$\ge2abc+3\sqrt[3]{(abc)^2}$
Hay $1\ge2t^3+3t^2\Leftrightarrow (2t-1)(t+1)^2\le0\Leftrightarrow t\le\frac{1}{2}$
Suy ra: $P=abc=t^3\le\frac{1}{8}$
Max$P=\frac{1}{8}\Leftrightarrow x=y=z=\frac{3}{2}$
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