\(B=a^3+b^3+ab=\left(a+b\right)\left(a^2-ab+b^2\right)+ab\)
Với \(a+b=1\)ta có: \(B=a^2-ab+b^2+ab=a^2+b^2\)\
Từ \(a+b=1\)\(\Rightarrow b=1-a\)
\(\Rightarrow B=a^2+\left(1-a\right)^2=a^2+1-2a+a^2=2a^2-2a+1\)
\(=2\left(a^2-a+\frac{1}{4}\right)+\frac{1}{2}=2\left(a^2-2.\frac{1}{2}a+\frac{1}{4}\right)+\frac{1}{2}\)
\(=2\left(a-\frac{1}{2}\right)^2+\frac{1}{2}\)
Vì \(\left(a-\frac{1}{2}\right)^2\ge0\forall a\)\(\Rightarrow2\left(a-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\forall a\)
hay \(B\ge\frac{1}{2}\)
Dấu " = " xảy ra \(\Leftrightarrow a-\frac{1}{2}=0\)\(\Leftrightarrow a=\frac{1}{2}\)
\(\Rightarrow b=1-\frac{1}{2}=\frac{1}{2}\)
Vậy \(minB=\frac{1}{2}\)\(\Leftrightarrow a=b=\frac{1}{2}\)
