Theo AM-GM ta có: \(VT=\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}\)
\(=\frac{a}{b}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}\)
\(\ge6\sqrt[6]{\frac{a}{b}\cdot\frac{1}{2}\sqrt{\frac{b}{c}}\cdot\frac{1}{2}\sqrt{\frac{b}{c}}\cdot\frac{1}{3}\sqrt[3]{\frac{c}{a}}\cdot\frac{1}{3}\sqrt[3]{\frac{c}{a}}\cdot\frac{1}{3}\sqrt[3]{\frac{c}{a}}}\)
\(=6\sqrt[6]{\frac{a}{b}\cdot\frac{1}{4}\cdot\frac{1}{27}\cdot\sqrt{\frac{b^2}{c^2}}\cdot\sqrt[3]{\frac{c^3}{a^3}}}\)
\(=6\sqrt[6]{\frac{a}{b}\cdot\frac{1}{108}\cdot\frac{b}{c}\cdot\frac{c}{a}}=6\sqrt[6]{\frac{1}{108}}=\frac{6}{\sqrt[6]{108}}>\frac{5}{2}\)
