Dan Sitaru's Cyclic Inequality with a Constraint and Cube Roots

Problem

Dan  Sitaru's Cyclic Inequality with a Constraint and Cube Roots

Solution 1

Observe that $(1+a)(1+b)(1+c)\ge (1+\sqrt[3]{abc})^3,$ implying

$\displaystyle \sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}\le\frac{\sqrt[3]{abc}}{1+\sqrt[3]{abc}}=1-\frac{1}{1+\sqrt[3]{abc}}.$

By Bergstrom's inequality, then

$\displaystyle \begin{align} \sum_{cycl}\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}&\le 4-\sum_{cycl}\frac{1}{1+\sqrt[3]{abc}}\\ &\le 4-\frac{16}{\displaystyle 4+\sum_{cycl}\sqrt[3]{abc}}\\ &\le 4-\frac{16}{5}=\frac{4}{5}. \end{align}$

Solution 2

Let, $\sqrt[3]{abc}=x.$

$\displaystyle \begin{align} &\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}=\sqrt[3]{\frac{1}{\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right) \left(1+\frac{1}{c}\right)} } \\ &=\sqrt[3]{\frac{1}{1+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) +\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+\frac{1}{abc}}} \\ &\leq \sqrt[3]{\frac{1}{1+\frac{3}{(abc)^{1/3}} +\frac{3}{(abc)^{2/3}}+\frac{1}{abc}}}~\text{(AM-GM)} \\ &=\sqrt[3]{\frac{1}{\left[1+\frac{1}{(abc)^{1/3}}\right]^3}}=\frac{x}{1+x}\sim\text{A concave function in $x$}. \end{align}$

Applying Jensen's,

$\displaystyle \begin{align} &LHS\leq\sum_{cycl}\frac{x}{1+x}\leq 4\left[\frac{\displaystyle \frac{1}{4}\sum_{cycl} x}{\displaystyle 1+\frac{1}{4}\sum_{cycl}x}\right] \\ &=\frac{4}{5}~\text{(Plugging in the constraint $\sum_{cycl} x =1$)}. \end{align}$

Acknowledgment

Dan Sitaru has kindly posted the problem at the CutTheKnotMath facebook page, with a solution by Nguyen Thanh Nho (Tri Nitrotoluen). The problem was originally publshed at the Romanian Mathematical Magazine. Solution 2 is by Amit Itagi.

 

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