Ta có 0 < y < x nên $x-y>0$
Xét hiệu ${{x}^{3}}-{{y}^{3}}={{x}^{3}}+{{x}^{2}}y+x{{y}^{2}}-{{x}^{2}}y-x{{y}^{2}}-{{y}^{3}}$
$=\left( {{x}^{3}}+{{x}^{2}}y+x{{y}^{2}} \right)-\left( {{x}^{2}}y+x{{y}^{2}}+{{y}^{3}} \right)$
$=x\left( {{x}^{2}}+xy+{{y}^{2}} \right)-y\left( {{x}^{2}}+xy+{{y}^{2}} \right)$
$=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)$
Ta thấy ${{x}^{2}}+xy+{{y}^{2}}={{x}^{2}}+2.x.\frac{1}{2}y+\frac{1}{4}{{y}^{2}}+\frac{3}{4}{{y}^{2}}={{\left( x+\frac{1}{2}y \right)}^{2}}+\frac{3}{4}{{y}^{2}}>0$
Mà $x-y>0$ nên $\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)>0$ hay ${{x}^{3}}-{{y}^{3}}>0\Rightarrow {{x}^{3}}>{{y}^{3}}$