solution $\frac{a}{b}+(\frac{\frac{b}{c}}{2}*2)^{\frac{1}{2}}+(\frac{\frac{c}{a}}{3}*3)^{\frac{1}{3}}\geq 6*(\frac{abc}{108abc})^{\frac{1}{6}}>2$ by A.M G.M

Doc.AK wrote: If $a,b,c$ are positive real numbers prove that: $\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2.$ It is not the question, but it is quite simple to find the exact minimum : $f(x)=\frac ax+\sqrt{\frac xc}$ has over $\mathbb R^+$ a minimum for $x=\sqrt[3]{4a^2c}$ and this minimum is $3\sqrt[3]{\frac a{4c}}$ Choosing $b=\sqrt[3]{4a^2c}$ required expression is, setting $y=\sqrt[3]{\frac ac}$ : $\frac 3{\sqrt[3]4}y+\frac 1y$ And minimum of this expression is easy to establish : $\boxed{2\frac{\sqrt 3}{\sqrt[3]2}>2}$

Doc.AK wrote: If $a,b,c$ are positive real numbers prove that: $\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2.$ True because: \[\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}=\frac{a}{b}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{2}\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}} \geq 3\sqrt[3]{\frac{a}{4c}} >\sqrt[3]{\frac{a}{c}} +\sqrt[3]{\frac{c}{a}} \geq 2\]

Doc.AK wrote: If $a,b,c$ are positive real numbers prove that: $\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2.$ See also here http://www.artofproblemsolving.com/community/c6h1244933p6380468

pco wrote: Doc.AK wrote: If $a,b,c$ are positive real numbers prove that: $\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2.$ It is not the question, but it is quite simple to find the exact minimum : $f(x)=\frac ax+\sqrt{\frac xc}$ has over $\mathbb R^+$ a minimum for $x=\sqrt[3]{4a^2c}$ and this minimum is $3\sqrt[3]{\frac a{4c}}$ Choosing $b=\sqrt[3]{4a^2c}$ required expression is, setting $y=\sqrt[3]{\frac ac}$ : $\frac 3{\sqrt[3]4}y+\frac 1y$ And minimum of this expression is easy to establish : $\boxed{2\frac{\sqrt 3}{\sqrt[3]2}>2}$ Why is f (x) equal to what you have written? Also, how did you find x (min) and then 3 square root...?

$f(x)$ is just a definition. So "why is f(x)..." has no meaning. It is just a definition. And $x_min$ is the unique positive value of $x$ where $f'(x)=0$.

Doc.AK wrote: If $a,b,c$ are positive real numbers prove that: $\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2.$ Let $a, b, c$ be positive real numbers. Prove that $$\frac {a} {b} + \sqrt [3]{\frac {b} {c}} + \sqrt [5] {\frac {c} {a}}>\frac{5}{2}.$$

Doc.AK wrote: If $a,b,c$ are positive real numbers prove that: $\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2.$ If $a,b,c$ are positive real numbers prove that: $$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$$

Doc.AK wrote: If $a,b,c$ are positive real numbers prove that: $\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2.$ Stop discussing problems for which the deadline hasn't passed yet. https://www.awesomemath.org/mathematical-reflections/current-issue/

winnertakeover wrote: Stop discussing problems for which the deadline hasn't passed yet. https://www.awesomemath.org/mathematical-reflections/current-issue/ Hemm, this thread has been created two years and six monthes ago ..... So it has no relation with the link you gave. I dont think we must demand AOPS to erase all existing problems just because a teacher / paper produces a public contest with old existing material on the web.

sqing wrote: Let $a, b, c$ be positive real numbers. Prove that $$\frac {a} {b} + \sqrt [3]{\frac {b} {c}} + \sqrt [5] {\frac {c} {a}}>\frac{5}{2}.$$ NaPrai wrote: For a better bound, \begin{align*}\sqrt{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[5]{\frac{c}{a}}&=10\left(\frac{2}{10}\cdot\left(\frac{1}{2}\sqrt{\frac{a}{b}}\right)+\frac{3}{10}\cdot\left(\frac{1}{3}\sqrt[3]{\frac{b}{c}}\right)+\frac{5}{10}\cdot\left(\frac{1}{5}\sqrt[5]{\frac{c}{a}}\right)\right)\\&\ge 10\left(\frac{1}{2}\sqrt{\frac{a}{b}}\right)^\frac{2}{10}\left(\frac{1}{3}\sqrt[3]{\frac{b}{c}}\right)^\frac{3}{10}\left(\frac{1}{5}\sqrt[5]{\frac{c}{a}}\right)^\frac{5}{10}\\&=\frac{10}{\sqrt[10]{337500}}\\&\sim 2.80009407285\end{align*}which equality for $a:b:c = 108:27k^2:k^5$, where $k=2^{0.2}3^{0.3}5^{0.5} \sim 3.57130858457$. https://artofproblemsolving.com/community/c6h1751059p11414969

sqing wrote: Let $a, b, c$ be positive real numbers. Prove that $$\frac {a} {b} + \sqrt [3]{\frac {b} {c}} + \sqrt [5] {\frac {c} {a}}>\frac{5}{2}.$$ Let $a,b,c >0$ and $a \geq b.$Then $$\frac{a}{b}+3\sqrt[3]{\frac{b}{c}+1}+5\sqrt[5]{\frac{c}{a}+1}>9\sqrt[9]{4}.$$$$\frac{a}{b}+2\sqrt{\frac{b}{c}+1}+3\sqrt[3]{\frac{c}{a}+1}>6\sqrt[3]{2}.$$here

Prove for any positives $a,b,c$ the inequality $$ \sqrt[3]{\dfrac{a}{b}}+\sqrt[5]{\dfrac{b}{c}}+\sqrt[7]{\dfrac{c}{a}}>\dfrac{5}{2}$$Kazakhstan 2019

sqing wrote: Doc.AK wrote: If $a,b,c$ are positive real numbers prove that: $\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2.$ If $a,b,c$ are positive real numbers prove that: $$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$$ By AM-GM and simple computation: $$\frac ab+\frac12\sqrt{\frac bc}+\frac12\sqrt{\frac bc}+\frac13\sqrt[3]{\frac ca}+\frac13\sqrt[3]{\frac ca}+\frac13\sqrt[3]{\frac ca}\ge6\sqrt[6]{\frac1{108}}>\frac52>2$$

$ \frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>\frac{1}{3}\frac{a}{b}+\frac{1}{3}\sqrt{\frac{b}{c}}+\frac{1}{3}\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}}\geq 6*\sqrt[6]{\frac{abc}{3^{6}*abc}}=2$ by AM GM