Answer
$f\left( {\frac{1}{3},\frac{1}{3},\frac{1}{3}} \right) = \frac{1}{3}$
Work Step by Step
$$\eqalign{
& {\text{Minimize }}f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2} \cr
& {\text{Constraint: }}x + y + z = 1{\text{ }} \to {\text{ }}g\left( x,y,z \right) = x + y + z \cr
& \cr
& {\text{We need to minimize }}f\left( x \right) = {x^2} + {y^2} + {z^2}{\text{ }}\left( {{\text{objective function}}} \right) \cr
& {\text{Using the Method of Lagrange Multipliers}} \cr
& *{\text{ }}{f_x}\left( {x,y,z} \right) = \lambda {g_x}\left( {x,y,z} \right) \cr
& \frac{\partial }{{\partial x}}\left[ {{x^2} + {y^2} + {z^2}} \right] = \lambda \frac{\partial }{{\partial x}}\left[ {x + y + z} \right] \cr
& 2x = \lambda \left( 1 \right) \cr
& 2x = \lambda {\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& *{\text{ }}{f_y}\left( {x,y,z} \right) = \lambda {g_y}\left( {x,y,z} \right) \cr
& \frac{\partial }{{\partial y}}\left[ {{x^2} + {y^2} + {z^2}} \right] = \lambda \frac{\partial }{{\partial y}}\left[ {x + y + z} \right] \cr
& 2y = \lambda \left( 1 \right) \cr
& 2y = \lambda {\text{ }}\left( {\bf{2}} \right) \cr
& \cr
& *{\text{ }}{f_z}\left( {x,y,z} \right) = \lambda {g_z}\left( {x,y,z} \right) \cr
& \frac{\partial }{{\partial z}}\left[ {{x^2} + {y^2} + {z^2}} \right] = \lambda \frac{\partial }{{\partial z}}\left[ {x + y + z} \right] \cr
& 2z = \lambda \left( 1 \right) \cr
& 2z = \lambda {\text{ }}\left( {\bf{3}} \right) \cr
& \cr
& {\text{Solving the system of equations }}\left( {\bf{1}} \right){\text{, }}\left( {\bf{2}} \right){\text{ and }}\left( {\bf{3}} \right) \cr
& {\text{Equating the equations }} \cr
& 2x = 2y = 2z \cr
& x = y = z \cr
& \cr
& {\text{Substituting }}y = x{\text{ and }}z = x{\text{ into the constraint }} \cr
& x + y + z = 1 = 0 \cr
& x + x + x = 1 \cr
& 3x = 1 \cr
& x = \frac{1}{3} \cr
& {\text{Then}}{\text{, }}y = \frac{1}{3}{\text{ and }}z = \frac{1}{3} \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& f\left( {\frac{1}{3},\frac{1}{3},\frac{1}{3}} \right) = {\left( {\frac{1}{3}} \right)^2} + {\left( {\frac{1}{3}} \right)^2} + {\left( {\frac{1}{3}} \right)^2} \cr
& f\left( {\frac{1}{3},\frac{1}{3},\frac{1}{3}} \right) = \frac{1}{3} \cr
& \cr
& {\text{The minimum is }}\frac{1}{3}{\text{ at }}x = y = z = \frac{1}{3} \cr} $$
