\(1-\frac{1}{1+a}+2-\frac{2}{1+b}=1\)
\(\Rightarrow\frac{1}{1+a}+\frac{2}{1+b}=1\)
\(\Rightarrow\frac{1}{1+a}=1-\frac{2}{1+b}=\frac{b-1}{b+1}\)
Do \(a>0\Rightarrow\frac{1}{1+a}>0\Rightarrow\frac{b-1}{b+1}>0\Rightarrow b>1\)
\(\Rightarrow a+1=\frac{b+1}{b-1}\Rightarrow a=\frac{b+1}{b-1}-1=\frac{2}{b-1}\)
\(\Rightarrow P=\frac{2b^2}{b-1}=2\left(b+1\right)+\frac{2}{b-1}=2\left(b-1\right)+\frac{2}{b-1}+4\)
\(\Rightarrow P\ge2\sqrt{\frac{4\left(b-1\right)}{b-1}}+4=8\)
Dấu "=" xảy ra khi \(a=b=2\)