Đặt [tex]a=x^2,b=y^2,c=z^2\Rightarrow xyz=1[/tex]
Ta có:[tex]P=\frac{1}{a+2b+3}+\frac{1}{b+2c+3}+\frac{1}{c+2a+3}=\frac{1}{x^2+2y^2+3}+\frac{1}{y^2+2z^2+3}+\frac{1}{z^2+2x^2+3}[/tex]
Vì [tex]\frac{1}{x^2+2y^2+3}=\frac{1}{x^2+y^2+y^2+1+2}\leq \frac{1}{2xy+2y+2}=\frac{1}{2}.\frac{1}{xy+y+1}[/tex]
[tex]\Rightarrow P\leq \frac{1}{2}(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{xz+x+1})=\frac{1}{2}(\frac{1}{xy+y+1}+\frac{xyz}{yz+z+xyz}+\frac{y}{xyz+xy+y})=\frac{1}{2}(\frac{1}{xy+y+1}+\frac{xy}{xy+y+1}+\frac{y}{xy+y+1})=\frac{1}{2}.1=\frac{1}{2}[/tex]
Dấu "=" xảy ra khi x = y = z = 1 hay a = b = c = 1.