find the minimum of $(a-b)^2+(a-1)^2+(b-1)^2$ Let $a,b,$ be real numbers such that $a\ge 4b$ . 

Find the minimum of $(a-b)^2+(a-1)^2+(b-1)^2$.
  This problem is  http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=553680


The minimum is either on the interior of the feasible set, or on the boundary.
The gradient of the objective is
$$\left[2(a-b)+2(a-1), -2(a-b)+2(b-1)\right] = \left[4a-2b-2, 4b-2a-2\right],$$
and so the unconstrained minimum is the infeasible point $(1,1)$. Therefore the minimum is on the boundary. Substituting $a=4b$ gives 
$$\min_b\ (3b)^2 + (4b-1)^2 + (b-1)^2 = \min_b\ 26b^2-10b+2.$$
The gradient is  $52b-10$, so $b=\frac{5}{26}$ and the minimum is $\frac{27}{26}$.
