# Hung Viet's Inequality III

### Statement

### Solution

Observe that by *Höder's inequality*

$\displaystyle \begin{align} &\left(\sum_{cycl}a^4\right)\left(\sum_{cycl}a^2b^2\right)\ge\left(\sum_{cycl}a^3b\right)^2,\\ &\left(\sum_{cycl}ab^3\right)\left(\sum_{cycl}a^3b\right)\ge\left(\sum_{cycl}a^2b^2\right)^2. \end{align}$

The product of the above two is exactly the required inequality.

### Acknowledgment

The inequality - by Nguyen Viet Hung - has been published at Spring issue of the Romanian Mathematical Magazine. This is Problem SP048. I reproduce here the charming solution by Kevin Soto Palacios. Additional solutions can be found at the link.

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