Đáp án:
Áp dụng BĐT ` Cô si ` ta có :
`a^3 + b^2 + c >= 3` $\sqrt[3]{a^3b^2c} = 3a\sqrt[3]{b^2c}$
`b^3 + c^2 + a >= 3` $\sqrt[3]{b^3c^2a} = 3b\sqrt[3]{c^2a}$
`c^3 + a^2 + b >= 3` $\sqrt[3]{c^3a^2b} = 3c\sqrt[3]{a^2b}$
`-> VT <= ` $\dfrac{a}{3a\sqrt[3]{b^2c}} + \dfrac{b}{3b\sqrt[3]{c^2a}} + \dfrac{c}{3c\sqrt[3]{a^2b}} = \dfrac{1}{3\sqrt[3]{b^2c}} + \dfrac{1}{3\sqrt[3]{c^2a}} + \dfrac{1}{3\sqrt[3]{a^2b}} = \dfrac{1}{9} . 3 . \sqrt[3]{\dfrac{1}{b} . \dfrac{1}{b} . \dfrac{1}{c}} + \dfrac{1}{9} . 3 . \sqrt[3]{\dfrac{1}{c} . \dfrac{1}{c} . \dfrac{1}{a}} + \dfrac{1}{9} . 3 . \sqrt[3]{\dfrac{1}{a} . \dfrac{1}{a} . \dfrac{1}{b}} \le \dfrac{1}{9}(\dfrac{1}{b} + \dfrac{1}{b} + \dfrac{1}{c}) + \dfrac{1}{9}(\dfrac{1}{c} + \dfrac{1}{c} + \dfrac{1}{a}) + \dfrac{1}{9}(\dfrac{1}{a} + \dfrac{1}{a} + \dfrac{1}{b}) = \dfrac{1}{3}(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}) = \dfrac{1}{3} . 3 = 1 = VP$
`-> đ.p.c.m`
Dấu "=" xảy ra `<=> a = b =c = 1`
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