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Inequality with roots

Let a, b, c be positive real numbers. Prove that

    \[\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2\]

Solution

We apply the AM – GM inequality, thus:

    \begin{align*} \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}} \\ &\geq 6\sqrt[6]{\frac{a}{b}\frac{b}{4c}\frac{c}{27a}} \\ &=\frac{6}{\sqrt[6]{4\cdot 27}} \end{align*}

Hence it suffices to prove that 3>\sqrt[6]{4\cdot 27} which holds because it is equivalent to 3^6> 4\cdot 27.

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