Thay y= 1-x ta được
\(c=x^2+y^2+xy=x^2+\left(1-x\right)^2+x\left(1-x\right)=x^2-x+1\)
\(=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Đẳng thức xảy ra \(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y=1-x\end{cases}}\) \(\Leftrightarrow x=y=\frac{1}{2}\)
Đặt \(x=1-y\)
\(C=x^2+y^2+xy=\left(1-y\right)^2+y^2+y\left(1-y\right)\)
\(\Leftrightarrow C=1-2y+y^2+y^2+y-y^2=y^2-y+1\)
\(\Leftrightarrow\left(y^2-2.\frac{1}{2}y+\frac{1}{4}\right)+\frac{3}{4}\Leftrightarrow\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy min C là 3/4 khi y=1/2 và x =1- 1/2= 1/2 hay x=y= 1/2
