Question

CMR: \(\dfrac{a}{b}+\sqrt{\dfrac{b}{c}}+\sqrt[3]{\dfrac{c}{a}}\ge\dfrac{5}{2}\)

giúp tớ với Nguyễn Thanh Hằng,nguyen van tuan,Nguyễn Huy Tú,Akai Haruma,Ace Legona

Answer 1

Đặt \(\dfrac{a}{b}=x;\sqrt{\dfrac{b}{c}}=y;\sqrt[3]{\dfrac{c}{a}}=z\)

\(\Rightarrow xy^2z^3=1\)

Ta cần chứng minh \(x+y+z\ge\dfrac{5}{2}\) (*)

Ta có \(x+y+z=x+\dfrac{y}{2}+\dfrac{y}{2}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\)

\(\ge6\sqrt[6]{x.\dfrac{y}{2}.\dfrac{y}{2}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=6\sqrt[6]{\dfrac{xy^2z^3}{108}}=6\sqrt[6]{\dfrac{1}{108}}>\dfrac{5}{2}\)

Như vậy (*) được chứng minh

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