Ta có:
\(\hept{\begin{cases}\frac{a^2}{1+b}+\frac{1+b}{4}\ge a\\\frac{b^2}{1+a}+\frac{1+a}{4}\ge b\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{a^2}{1+b}\ge\frac{4a-b-1}{4}\\\frac{b^2}{1+a}\ge\frac{4b-a-1}{4}\end{cases}}\)
\(\Rightarrow A=\frac{a^2}{1+b}+\frac{b^2}{1+a}\ge\frac{4a-b-1}{4}+\frac{4b-a-1}{4}\)
\(=\frac{3}{4}\left(a+b\right)-\frac{1}{2}\ge\frac{3}{4}.2\sqrt{ab}-\frac{1}{2}=\frac{3}{2}-\frac{1}{2}=1\)
Dấu = xảy ra khi \(a=b=1\)