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vansang02121998

$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$

$\Leftrightarrow \dfrac{xy+xz+yz}{xyz}=1$

$\Leftrightarrow xy+yz+zx=xyz$

Lại có

$(a-1)(b-1)(c-1)=abc-ab-ac-bc+a+b+c-1$

$\Leftrightarrow (a-1)(b-1)(c-1)=a+b+c-1$

Áp dụng Cauchy

$(x+y+z)(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}) \ge 9$

$\Leftrightarrow x+y+z \ge 9$

$\Leftrightarrow x+y+z-1 \ge 8$

Vậy, $(a-1)(b-1)(c-1) \ge 8$

Dấu $"="$ xảy ra khi $a=b=c=3$
 
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