Inequality $\left(1-\frac{x(1-x)+y(1-y)}{1-x+1-y}\right)^2+(1-\frac{x+y}{2})^2\geq (1-x)^2+(1-y)^2$

For $x+y\neq2$ by AM-GM we obtain:$$\left(1-\frac{x(1-x)+y(1-y)}{1-x+1-y}\right)^2+\left(1-\frac{x+y}{2}\right)^2=$$ $$=\left(\frac{(1-x)^2+(1-y)^2}{2-x-y}\right)^2+\left(\frac{2-x-y}{2}\right)^2\geq$$ $$\geq2\sqrt{\left(\frac{(1-x)^2+(1-y)^2}{2-x-y}\right)^2\left(\frac{2-x-y}{2}\right)^2}= (1-x)^2+(1-y)^2$$