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Câu hỏi: Cho ba số dương a, b, c, chứng minh rằng :
\(\left( {1 + \dfrac{a}{b}} \right)\left({{\rm{a}} + \dfrac{b}{c}} \right)\left({1 + \dfrac{c}{a}} \right) \ge 8\)
\(\left( {1 + \dfrac{a}{b}} \right)\left({{\rm{a}} + \dfrac{b}{c}} \right)\left({1 + \dfrac{c}{a}} \right) \ge 8\)
Lời giải chi tiết
Với \(a > 0, b > 0, c > 0\) thì
\(1 + \dfrac{a}{b} \ge 2\sqrt {\dfrac{a}{b}} \ge 0;\)
\(1 + \dfrac{b}{c} \ge 2\sqrt {\dfrac{b}{c}} ;\)
\(1 + \dfrac{c}{a} \ge 2\sqrt {\dfrac{c}{a}} \ge 0\)
Từ đó suy ra \(\left( {1 + \dfrac{a}{b}} \right) \left({1 + \dfrac{b}{c}} \right)\left({1 + \dfrac{c}{a}} \right) \ge {2^3}\sqrt {\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{a}}\)
\( = 8\)
Với \(a > 0, b > 0, c > 0\) thì
\(1 + \dfrac{a}{b} \ge 2\sqrt {\dfrac{a}{b}} \ge 0;\)
\(1 + \dfrac{b}{c} \ge 2\sqrt {\dfrac{b}{c}} ;\)
\(1 + \dfrac{c}{a} \ge 2\sqrt {\dfrac{c}{a}} \ge 0\)
Từ đó suy ra \(\left( {1 + \dfrac{a}{b}} \right) \left({1 + \dfrac{b}{c}} \right)\left({1 + \dfrac{c}{a}} \right) \ge {2^3}\sqrt {\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{a}}\)
\( = 8\)