\(A=a\left(a^2+2b\right)+b\left(b^2-a\right)\)

\(=a^3+2ab+b^3-ab=a^3+b^3+ab\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+ab\)

\(=a^2-ab+b^2+ab\left(a+b=1\right)\)

\(=a^2+b^2\). Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(1^2+1^2\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2=1\)

\(\Rightarrow2\left(x^2+y^2\right)\ge1\Rightarrow x^2+y^2\ge\dfrac{1}{2}\Rightarrow A\ge\dfrac{1}{2}\)

Đẳng thức xảy ra khi \(a=b=\dfrac{1}{2}\)