ta có :
\(A=\frac{a^2}{1-a}+a+\frac{b^2}{1-b}+b+\frac{1}{a+b}=\frac{a}{1-a}+\frac{b}{1-b}+\frac{1}{a+b}\)
\(A=\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{a+b}-2\)
mà : \(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{a+b}\ge\frac{9}{1-a+1-b+a+b}=\frac{9}{2}\)
Vậy \(A\ge\frac{9}{2}-2=\frac{5}{2}\)
dấu bằng xảy ra khi : \(1-a=1-b=a+b\Leftrightarrow a=b=\frac{1}{3}\)