# Hung Viet's Inequality II

### Solution

$\displaystyle \frac{1}{5}\sum_{cycl}a^2=\frac{1}{20}\left[\left(\sum_{cycl}a\right)^2+\sum_{all}(a-b)^2\right],$

implying,

$\displaystyle \frac{1}{20}\sum_{all}(a-b)^2\le \frac{1}{5}\sum_{cycl}a^2.$

Assume, WLOG, that $a\le b\le c\le d.\,$ Then

$d-a\ge d-b\ge d-c\,$ and $c-a\ge c-b.$

It follows that

$\displaystyle \min_{all}(a-b)=\min (d-c,c-b,b-a)$

Let $b-a=x,\,$ $c-b=y,\,$ and $d-c=z.\,$ Then $x,y,z\ge 0\,$ and $d-a=x+y+z,\,$ $d-b=y+z,\,$ $c-a=x+y.\,$ Also, the minimum in question is one of $x^2,y^2,z^2.\,$ So we need to prove

$\displaystyle \min \{x^2,y^2,z^2\}\le\frac{(x+y+z)^2+(x+y)^2+(y+z)^2+x^2+y^2+z^2}{20},$

which is obvious. Equality as attained at $x=y=z\,$ and $a+b+c+d=0,\,$ or, equivalently, for $(a,b,c,d)=(-3t,-t,t,3t),\,$ for $t\in\mathbb{R}\,$ and permutations.

### Acknowledgment

The inequality and the solution have been shared at the CutTheKnotMath facebook page by Leo Giugiuc, with a link to a post by Nguyen Viet Hung at the mathematical inequalities facebook group where it appeared originally.