# An Inequality in Integers II

### Solution

WLOG, $a\gt b\gt c.\;$ We need to prove $\displaystyle\frac{1}{6}(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]\ge 3k.\;$

But $a-b\ge 1,\;$ $b-c\ge 1,\;$ $a-c\ge 2\;$ so that $\displaystyle\frac{1}{6}[(a-b)^2+(b-c)^2+(c-a)^2]\ge 1.\;$ Hence, suffice it to prove that $a+b+c\ge 3k.$

On the other hand, since $a,b,c\;$ are distinct, $a+b+c\gt\sqrt{3(ab+bc+ca)},\;$ implying $a+b+c\gt\sqrt{9k^2-3}.\;$ But $a+b+c\;$ is a positive integer, so $a+b+c\ge 3k.$

### Acknowledgment

The problem, along with the solution, has been posted by Leo Giugiuc at CutThKnotMath facebook page.

[an error occurred while processing this directive]