# Hung Viet's Inequality

### Solution

The required inequality is equivalent to

$\displaystyle 2\sum_{cycl}a^3+\sum_{cycl}(a^3+b^3)\sum_{cycl}\frac{ab(a+b)}{a^3+b^3}\ge 3\sum_{cycl}a\cdot\sum_{cycl}ab-3abc.$

By the Cauchy-Schwarz (or Bergström's) inequality,

$\displaystyle \sum_{cycl}(a^3+b^3)\sum_{cycl}\frac{ab(a+b)}{a^3+b^3}\ge \left[\sum_{cycl}\sqrt{ab(a+b)}\right]^2.$

Thus, suffice it to show that

$\displaystyle 2\sum_{cycl}a^3+\left[\sum_{cycl}\sqrt{ab(a+b)}\right]^2\ge 3\sum_{cycl}a\cdot\sum_{cycl}ab-3abc$

which comes to

$\displaystyle\sum_{cycl}a^3+\sqrt{abc}\sum_{cycl}\sqrt{a(a+b)(a+c)}\ge\sum_{cycl}a\cdot\sum_{cycl}ab.$

This is equivalent to

$\displaystyle\sum_{cycl}a^3+\sqrt{abc}\sum_{cycl}\sqrt{a(a+b)(a+c)}\ge \sum_{cycl}a\cdot\sum_{cycl}ab.$

By the AM-GM inequality, $\displaystyle\sum_{cycl}\sqrt{a(a+b)(a+c)}\ge 6\sqrt{abc}.\;$ So, the required inequality will follow from

$a^3+b^3+c^3+6abc\ge (a+b+c)(ab+bc+ca),$

but this is Schur's inequality with $t=1.$

### Acknowledgment

The inequality and the solution have been shared at the CutTheKnotMath facebook page by Leo Giugiuc who credits the inequality to Nguyen Viet Hung.