Lolita_(✿˵◕ ɜ◕˵)
[imath]E=\left(1+\dfrac{b}{3c}+\dfrac{a}{3b}+\dfrac{a}{9c}\right)\left(1+\dfrac{c}{3a}\right)[/imath]
[imath]=1+\dfrac{c}{3a}+\dfrac{b}{3c}+\dfrac{b}{9a}+\dfrac{a}{3b}+\dfrac{c}{9b}+\dfrac{a}{9c}+\dfrac{1}{27}[/imath]
[imath]=\dfrac{28}{27}+\dfrac{1}{3}\left(\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{a}{b}\right)+\dfrac{1}{9}\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)[/imath]
[imath]\ge\dfrac{28}{27}+\sqrt[3]{\dfrac{c}{a}.\dfrac{b}{c}.\dfrac{a}{b}}+\dfrac{1}{3}\sqrt[3]{\dfrac{b}{a}.\dfrac{c}{b}.\dfrac{a}{c}}=\dfrac{64}{27}[/imath]
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[Lý thuyết] Chuyên đề HSG : Bất đẳng thức