An Inequality in Reciprocals III


An  Inequality  in Reciprocals III


First we'll prove


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.


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.


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