Question

Cho a, b, c \(\ne\)0 thỏa mãn: abc=27 và \(a+b+c=\frac{9}{a}+\frac{9}{b}+\frac{9}{c}.\)

Chứng minh: ( a - 3 )( b - 3 )( c - 3 ) = 0.

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Answer 1

\(a+b+c=\frac{9}{a}+\frac{9}{b}+\frac{9}{c}=9\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=9\)

Ta có: \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1+1+1+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge3+2+2+2=9\)

Dấu "=" xảy ra khi a = b = c = 3 (do abc = 27)

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