Question

Đã trả lời bởi chuyên gia

Answer 1

Ta có:

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

\( = 3 + \left( {\frac{a}{b} + \frac{b}{a}} \right) + \left( {\frac{c}{b} + \frac{b}{c}} \right) + \left( {\frac{a}{c} + \frac{c}{a}} \right)\)

Áp dụng bất đẳng thức Cô – si ta có

\(\frac{a}{b} + \frac{b}{a} \ge 2\sqrt {\frac{a}{b}.\frac{b}{a}} = 2\)

\(\frac{c}{b} + \frac{b}{c} \ge 2\sqrt {\frac{c}{b}.\frac{b}{c}} = 2\)

\(\frac{a}{c} + \frac{c}{a} \ge 2\sqrt {\frac{a}{c}.\frac{c}{a}} = 2\)

Suy ra \(\left( {a + b + c} \right)\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) \ge 3 + 2 + 2 = 9\)

Vậy \(\left( {a + b + c} \right)\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) \ge 9\).

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Answer 2