Question 1016895
{{{a^2 + b^2  = a^2 + (1-a)^2 = 2a^2 -2a+1}}}
Now for any quadratic expression {{{Ax^2 +Bx + C}}}, the maximum or minimum value is {{{C-B^2/(4AC)}}}.  In this case, A = 2 >0, and hence we have a minimum value.

The minimum value of {{{2a^2 -2a+1}}} is then equal to {{{1-(-2)^2/(4*2*1) = 1 - 4/8 = 1/2}}}.

This ends the solution.