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Fie \( a\in(-2,2) \) , \( b\in(-3,3) \) , \( c\in(-4,4) \) astfel incat \( S=\frac{1}{2-a}+\frac{1}{3-b}+\frac{1}{4-c}=\frac{1}{2+a}+\frac{1}{3+b}+\frac{1}{4+c} \)

Demonstrati ca \( S\geq1 \)

Avem \( 2S=\frac{1}{2-a}+\frac{1}{2+a}+\frac{1}{3-b}+\frac{1}{3+b}+\frac{1}{4-c}+\frac{1}{4+c}\ge \frac{36}{2-a+2+a+3-b+3+b+4-c+4+c}=\frac{36}{18}=2\ \ (Cauchy) \)
Deci, \( 2S\ge 2\Rightarrow S\ge 1 \).