\(M=x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^2-xy+y^2=\left(1-y\right)^2-\left(1-y\right)y+y^2\)
\(=3y^2-3y+1=\left(3y^2-3y+\frac{3}{4}\right)+\frac{1}{4}\)
\(=3\left(y-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)
Vậy GTNN là \(\frac{1}{4}\) dạt được khi \(x=y=\frac{1}{2}\)