a+b=1
=>a=1-b
\(A=a\left(a^2+2b\right)+b\left(b^2-a\right)\)
\(=a^3+2ab+b^3-ab\)
\(=a^3+b^3+ab=\left(a+b\right)^3-3ab\left(a+b\right)+ab\)
\(=1^3-3ab+ab=1-2ab\)
=1-2b(1-b)
\(=1-2b+2b^2=2\left(b^2-b+\frac12\right)=2\left(b^2-b+\frac14+\frac14\right)\)
\(=2\cdot\left(b-\frac12\right)^2+\frac12\ge\frac12\forall b\)
Dấu '=' xảy ra khi \(b-\frac12=0\)
=>\(b=\frac12\)
=>\(a=1-b=1-\frac12=\frac12\)