The Lagrangian is
[tex]L(x,y,\lambda) = x^2 + y^2 - \lambda (x + y - 1)[/tex]
Find its critical points.
[tex]\dfrac{\partial L}{\partial x} = 2x - \lambda = 0 \implies x = \dfrac\lambda2[/tex]
[tex]\dfrac{\partial L}{\partial y} = 2y - \lambda = 0 \implies y = \dfrac\lambda2[/tex]
[tex]\dfrac{\partial L}{\partial\lambda} = -(x+y-1) = 0 \implies x + y = 1[/tex]
From the first two conditions, [tex]x=y[/tex], so
[tex]x + y = 2x = 1 \implies x = y = \dfrac12[/tex]
At the point (1/2, 1/2), compute the Hessian determinant of [tex]f(x,y)[/tex].
[tex]H(x,y) = \begin{bmatrix} \dfrac{\partial^2f}{\partial x^2} & \dfrac{\partial^2f}{\partial x \partial y} \\\\ \dfrac{\partial^2f}{\partial y\partial x} & \dfrac{\partial^2f}{\partial y^2} \end{bmatrix} = \begin{bmatrix}2&0\\0&2\end{bmatrix}[/tex]
[tex]H(x,y)[/tex] is positive definite for all [tex]x,y[/tex], which indeed indicates a local minimum at (1/2, 1/2), so that
[tex]\min\left\{x^2+y^2 \mid x+y=1\right\} = f\left(\dfrac12,\dfrac12\right) = \dfrac14+\dfrac14 = \boxed{\dfrac12}[/tex]