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An inequality

By Tolaso in Uncategorized on January 23, 2017.

Let $a, b, c$ be positive real numbers such that

$a+b+c=1$

Prove that

$\displaystyle \prod_{\text{cyclic}}\left ( \frac{1}{a} + \frac{1}{bc} \right ) > 1728$

(Vojtech Jarnik / Second Category / 2016)

Solution

Using the AM – GM inequality we have that

$\begin{align*} \frac{1}{a} + \frac{1}{bc}&= \frac{1}{a} + \frac{1}{3bc} + \frac{1}{3bc} + \frac{1}{3bc} \\ &> \frac{4}{\sqrt[4]{27 ab^3 c^3}} \end{align*}$

as well as $\displaystyle \left ( \frac{a+b+c}{3} \right )^3 = \frac{1}{27} >abc$ . Thus:

$\begin{align*} \prod_{\text{cyclic}} \left ( \frac{1}{a} + \frac{1}{bc} \right ) &> 64 \prod_{\text{cyclic}} \frac{1}{\sqrt[4]{27ab^3c^3}} \\ &=\frac{64}{\sqrt[4]{3^9 (abc)^7}} \\ &> \frac{64}{\sqrt[4]{3^9 \left ( 3^{-3} \right )^7}}\\ &= 64 \cdot \sqrt[4]{3^{12}} \\ &= 1728 \end{align*}$