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Fie \( x,y,z>0 \) astfel incat \( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 \) . Sa se demonstreze ca :

\( (x-1)(y-1)(z-1)\ge 8 \)

Din enunt obtinem: \( xy+yz+zx=xyz \)

Se cunoaste inegalitatea: \( \left\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right\)(x+y+z)\geq 9. \)

\( \Longrightarrow x+y+z\geq 9 \)

Inegalitatea de demonstrat devine: \( xyz-xy-yz-zx+x+y+z\geq 9 \)

\( \Longrightarrow x+y+z\geq 9 \), adevarat.

Fie \( x=\frac{a+b+c}{a},\ y=\frac{a+b+c}{b},\ z=\frac{a+b+c}{c}\ \left(a,\ b,\ c>0\right) \). Atunci \( \prod\left(x-1\right)=\prod\frac{b+c}{a}=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\ge\frac{8\sqrt{ab\cdot bc\cdot ca}}{abc}=8 \)