Chứng minh bất đẳng thức sau với $x,\,y> 0$$:$
$$\sqrt{\frac{x^{\,2}+ xy+ y^{\,2}}{3\,x^{\,2}- 3\,xy+ 3\,y^{\,2}}}\geqq \frac{x^{\,2}+ 2\,xy- y^{\,2}}{2\,x^{\,2}}$$
$<$$=$$>$
$\frac{4\,x^{\,4}(\,x^{\,2}+ xy+ y^{\,2}\,)- 3\,(\,x^{\,2}- xy+ y^{\,2}\,)(\,x^{\,2}+ 2\,xy- y^{\,2}\,)^{\,2}}{(\,x- y\,)^{\,2}}=$
$= x^{\,4}- 3\,x^{\,3}y+ 9\,xy^{\,3}- 3\,y^{\,4}=$
$= \frac{(\,x^{\,2}+ 2\,xy- y^{\,2}\,)(\,x^{\,2}- 5\,xy+ 7\,y^{\,2})+ x^{\,4}- 3\,x^{\,3}y+ 4\,x^{\,2}y^{\,2}- xy^{\,3}+ y^{\,4}}{2}$
