We expand the original expression, then factor the result by grouping: $$ \begin{aligned} (x y-1)^{2}+(x+y)^{2} \&=\left(x^{2} y^{2}-2 x y+1\right)+\left(x^{2}+2 x y+y^{2}\right) \textbackslash\textbackslash{} \&=x^{2} y^{2}+x^{2}+y^{2}+1 \textbackslash\textbackslash{} \&=x^{2}\left(y^{2}+1\right)+\left(y^{2}+1\right) \textbackslash\textbackslash{} \&=\left(x^{2}+1\right)\left(y^{2}+1\right) . \end{aligned} $$ Clearly, both factors are positive. By the Trivial Inequality, we have $$ \left(x^{2}+1\right)\left(y^{2}+1\right) \geq(0+1)(0+1)= 1 . $$ Note that the least possible value of $(x y-1)^{2}+(x+y)^{2}$ occurs at $x=y=0$. 
What is the least possible value of $(x y-1)^{2}+(x+y)^{2}$ for real numbers $x$ and $y$? 
Options:
[a] $$0$$ 
[b] $\frac{1}{4}$ 
[c] $\frac{1}{2}$ 
[d] $$1$$ 
[e] $$2$$ 