The following inequality is also true.

Let $a, b, c$ be positive real numbers. Prove that $$\frac {a} {b} + \sqrt [3]{\frac {b} {c}} + \sqrt [5] {\frac {c} {a}}>\frac{5}{2}.$$Maybe $$ \sqrt[5]{\frac {a} {b} }+ \sqrt [7]{\frac {b} {c}} + \sqrt [9] {\frac {c} {a}}>\frac{5}{2}.$$

By AM-GM \begin{align*} \sqrt[3]{\frac{a}{b}}+\sqrt[5]{\frac{b}{c}}+\sqrt[7]{\frac{c}{a}}&=\frac{1}{5}\cdot 5\sqrt[3]{\frac{a}{b}}+\frac{1}{3}\cdot 3\sqrt[5]{\frac{b}{c}}+\frac{7}{15}\cdot \frac{15}{7}\sqrt[7]{\frac{c}{a}} \\ &\geqslant \left(5\sqrt[3]{\frac{a}{b}}\right)^\frac{1}{5}\left(3\sqrt[5]{\frac{b}{c}}\right)^\frac{1}{3}\left(\frac{15}{7}\sqrt[7]{\frac{c}{a}}\right)^\frac{7}{15} \\ &= \frac{3^\frac{4}{5}\cdot 5^\frac{2}{3}}{7^\frac{7}{15}} \\ &> \frac{5}{2}. \end{align*}

DerJan wrote: By AM-GM \begin{align*} \sqrt[3]{\frac{a}{b}}+\sqrt[5]{\frac{b}{c}}+\sqrt[7]{\frac{c}{a}}&=\frac{1}{5}\cdot 5\sqrt[3]{\frac{a}{b}}+\frac{1}{3}\cdot 3\sqrt[5]{\frac{b}{c}}+\frac{7}{15}\cdot \frac{15}{7}\sqrt[7]{\frac{c}{a}} \\ &\geqslant \left(5\sqrt[3]{\frac{a}{b}}\right)^\frac{1}{5}\left(3\sqrt[5]{\frac{b}{c}}\right)^\frac{1}{3}\left(\frac{15}{7}\sqrt[7]{\frac{c}{a}}\right)^\frac{7}{15} \\ &= \frac{3^\frac{4}{5}\cdot 5^\frac{2}{3}}{7^\frac{7}{15}} \\ &> \frac{5}{2}. \end{align*} Splendid. Proof of Zhangyunhua:

Attachments:

sqing wrote: Let $a, b, c$ be positive real numbers. Prove that $$\frac {a} {b} + \sqrt [3]{\frac {b} {c}} + \sqrt [5] {\frac {c} {a}}>\frac{5}{2}.$$ https://artofproblemsolving.com/community/c6h1250316p11332607 rightways wrote: Prove for any positives $a,b,c$ the inequality $$ \sqrt[3]{\dfrac{a}{b}}+\sqrt[5]{\dfrac{b}{c}}+\sqrt[7]{\dfrac{c}{a}}>\dfrac{5}{2}$$ Let $ n\in N^+.$ Prove that $$\sqrt[n]{\dfrac{a}{b}}+\sqrt[n+2]{\dfrac{b}{c}}+\sqrt[n+4]{\dfrac{c}{a}}>\dfrac{5}{2}$$(Fire dance phoenix)

Let $a,b,c$ be positive reals . Prove that$$\sqrt{\dfrac ab} + \sqrt[3]{\dfrac bc} + \sqrt[5]{\dfrac ca} > \dfrac{14}{5}$$