Ta có: $\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}=\frac{a}{b}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}\ge 6\sqrt[6]{\frac{a}{b}.\frac{1}{2}\sqrt{\frac{b}{c}}.\frac{1}{2}\sqrt{\frac{b}{c}}.\frac{1}{3}\sqrt[3]{\frac{c}{a}}.\frac{1}{3}\sqrt[3]{\frac{c}{a}}.\frac{1}{3}\sqrt[3]{\frac{c}{a}}}=6\sqrt[6]{\frac{1}{2^2}.\frac{1}{3^3}}=6\sqrt[6]{\frac{1}{108}}>\frac{5}{2}$.
Vậy ta có điều phải chứng minh.
Woww! Em cảm ơn anh nhiều. Dễ vậy mà nhìn không ra