x + y = 1 <=> y = 1 - x

We have: F = x² + (1 - x)² + (1 - x).x

                   = x² + 1 + x² - 2x + x - x²

                   = x² - x + 1

                   \(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Equal signs occur when and only of \(x-\dfrac{1}{2}=0\) \(\Leftrightarrow x=\dfrac{1}{2}\)

\(y=1-\dfrac{1}{2}=\dfrac{1}{2}\)

min A \(=\dfrac{1}{2}\Leftrightarrow x=y=\dfrac{1}{2}\)

x+y=1             \(\Rightarrow\)\(x^2+2xy+y^2=1\)\(\left(1\right)\)

which \(\left(x-y\right)^2\ge0\)     else \(x^2-2xy+y^2\ge0\left(2\right)\)

plus 1 and 2 we have   \(2\left(x^2+y^2\right)\ge1\Rightarrow x^2+y^2\ge\dfrac{1}{2}\)

minA=\(\dfrac{1}{2}\)      When and only when  x = y = \(\dfrac{1}{2}\)