# An Inequality in Reciprocals III

### Solution

First we'll prove

Lemma

For positive $x,y,z,$

$\displaystyle\sum_{cycl}x^3y^3+xyz\sum_{cycl}x^3\ge xyz\sum_{cycl}xy(x+y).$

To prove that observe that, by the AM-GM inequality, $\displaystyle xyz\sum_{cycl}x^3\ge 3x^2y^2z^2.\,$ Thus suffice it to prove

$\displaystyle\sum_{cycl}x^3y^3+3x^2y^2z^2\ge xyz\sum_{cycl}xy(x+y).$

But this is a special case of Schur's inequality for the positive numbers $xy,\,$ $yz,\,$ $zx,\,$ with $t=1.$

Now, letting $x=t^a,\,$ $y=t^b,\,$ $z=t^c,\,$ $t\in [0,1],\,$ we have

$\displaystyle\sum_{cycl}t^{3(a+b)}+\sum_{cycl}t^{4a+b+c}\ge \sum_{cycl}\left(t^{a+2b+3c}+t^{a+3b+2c}\right).$

It follows that

$\displaystyle\sum_{cycl}\int_0^1\frac{1}{t}\cdot t^{3(a+b)}dt+\sum_{cycl}\int_0^1\frac{1}{t}\cdot t^{4a+b+c}dt\ge \sum_{cycl}\int_0^1\frac{1}{t}\left(t^{a+2b+3c}+t^{a+3b+2c}\right)dt.$

Simple integration completes the proof.

### Acknowledgment

The inequality that is an invention of Hung Nguyen Viet (Vietnam), has been posted at the CutTheKnotMath facebook page, along with his solution, by Leo Giugiuc.