Amateur of Mathematics

$P=\sum\frac{1}{x^2+2y^2+3}$
$\Leftrightarrow P=\sum \frac{1}{x^{2}+y^{2}+y^{2}+1+2}\leq \sum \frac{1}{2xy+2y+2}\leq \frac{1}{2}\sum \frac{1}{xy+y+1}$$\doteq \frac{1}{2}$
vì $\sum \frac{1}{xy+y+1}=1$
dấu"=" khi x=y=z=1
p/s:BĐT quá easy!!!!!!!!!!!!!