cho x,y,z dương thoả mãn x+y+z=6. CMR $P=x^{2}+y^{2}+z^{2}-xy-zx-zy+xyz\geq 8$
Áp dụng BĐT Am-Gm
Ta có : $xyz\geq \left ( x+y-z \right )\left ( x+z-y \right )\left ( z+x-y \right )=8\prod \left ( 3-x \right )=216+24\left ( xy+yz+zx \right )-8xyz-72\left ( x+y+z \right )\Rightarrow 9xyz\geq 24\left ( xy+yz+xz \right )-216\Rightarrow xyz\geq \frac{8}{3}(xy+yz+xz)-24$
$\Rightarrow P\geq \sum x^{2}-\sum xy+\frac{8}{3}\sum xy-24\geq \left ( x+y+z \right )^{2}-3\sum xy+\frac{8}{3}\sum xy-24=12-\frac{1}{3}\sum xy\geq 12-\frac{1}{9}\left ( x+y+z \right )^{2}=8$