|
|
Đặt $a=4x^2,b=4y^2,c=4z^2$ thì $xyz=1$.
Ta có:
$x^2+2y^2+3=x^2+y^2+y^2+1+2\geq 2xy+2y+2=2(xy+y+1).$
$\Rightarrow P=\frac{1}{4}\sum_{cyc}{\frac{1}{x^2+2y^2+3}}\geq \frac{1}{8}\sum_{cyc}{\frac{1}{1+y+xy}}=\frac{1}{8}$.
Dấu $=$ xảy ra khi vả chỉ khi $a=b=c=4$.
|