# An Inequality with Constraint in Four Variables V

### Statement

### Solution 1

By *Hölder's inequality*,

$\displaystyle \begin{align} &(1+a^3)(1+b^3)(1+c^3)\ge (1+abc)^3\\ &\sqrt[3]{(1+a^3)(1+b^3)(1+c^3)}\geq 1+abc\\ &\frac{1}{\sqrt[3]{(1+a^3)(1+b^3)(1+c^3)}}\leq \frac{1}{1+abc} \end{align}$

It follows that

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

$\displaystyle \sum_{cycl}\frac{1}{1+abc} \ge \frac{\displaystyle (1+1+1+1)^2}{\displaystyle 4+\sum_{cycl}abc}=\frac{16}{5}.$

Thus, $\displaystyle -\sum_{cycl}\frac{1}{1+abc}\leq -\frac{16}{5}.\,$ Further,

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

### Solution 2

By *Hölder's inequality*, $\displaystyle \sqrt[3]{\prod_{cycl}(1+a^3)}\ge 1+abc,\,$ implying

$\displaystyle 5\sum_{cycl}\frac{\displaystyle abc}{\displaystyle \sqrt[3]{\prod_{cycl}(1+a^3)}}\le5\sum_{cycl}\frac{abc}{1+abc}.$

By Jensen's inequality applied to the concave function $\displaystyle f(x)=\frac{x}{1+x},\,$

$\displaystyle 5\sum_{cycl}\frac{abc}{1+abc}\le 20\frac{frac{1}{4}}{\frac{5}{4}}=4.$

### Acknowledgment

Dan Sitaru has kindly posted a problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page and later emailed his solution in a LaTeX file. Solution 2 is by Leo Giugiuc. Arben Ajredini has independently came up with Solution 1.

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