This is exercise 3.3.1 in the book Nonlinear Programming by Dimitri Bertsekas:
Solve the problem
$$\min \hspace{0.2cm}(x-a)^2+(y-b)^2+xy \quad \text{s.t.} \quad 0\leq x,y\leq 1$$ for all possible values of $a$ and $b$.
I started by writing the KKT necessary conditions for this problem (LICQ condition is statisfied):
$$2(x-a)+y -\mu_1+\mu_3=0,$$
$$ 2(y-b)+x-\mu_2+\mu_4=0,$$
$$\mu_1x=0, \quad \mu_2y=0,$$
$$\mu_3(x-1)=0, \quad \mu_4(y-1)=0,$$
$$0\leq x,y\leq 1,$$
where $\mu_1,\mu_2,\mu_3,\mu_4\geq 0$. This leads to nine cases:
Case 1: $\mu_1=\mu_2=\mu_3=\mu_4=0$. Then $x=\frac{4a-2b}{3}$ and $y=\frac{4b-2a}{3}$, and the parameter restrictions are $0\leq 2a-b\leq 3/2$ and $0\leq 2b-a\leq 3/2$.
Case 2: $\mu_1>0,\mu_2=\mu_3=\mu_4=0$. Then $x=0$ and $y=b$, and the parameter restrictions are $b-2a>0$ and $1\leq b\leq 1$.
Case 3: $\mu_2>0,\mu_1=\mu_3=\mu_4=0$. Then $x=a$ and $y=0$, and the parameter restrictions are $a-2b>0$ and $1\leq a\leq 1$.
Case 4: $\mu_3>0,\mu_1=\mu_2=\mu_4=0$. Then $x=1$ and $y=b-1/2$, and the parameter restrictions are $2a-b>3/2$ and $1/2\leq b\leq 3/2$.
Case 5: $\mu_4>0,\mu_1=\mu_2=\mu_3=0$. Then $x=a-1/2$ and $y=1$, and the parameter restrictions are $2b-a>3/2$ and $1/2\leq a\leq 3/2$.
Case 6: $\mu_1,\mu_2>0,\mu_3=\mu_4=0$. Then $x=0$ and $y=0$, and the parameter restrictions are $a<0$ and $b<0$.
Case 7: $\mu_1,\mu_4>0,\mu_2=\mu_3=0$. Then $x=0$ and $y=1$, and the parameter restrictions are $a<1/2$ and $b>1$.
Case 8: $\mu_2,\mu_3>0,\mu_1=\mu_4=0$. Then $x=1$ and $y=0$, and the parameter restrictions are $b<1/2$ and $a>1$.
Case 9: $\mu_3,\mu_4>0,\mu_1=\mu_2=0$. Then $x=1$ and $y=1$, and the parameter restrictions are $a>3/2$ and $b>3/2$.
Inspection shows that the nine parameter restrictions are mutually exclusive and exhaustive. Hence all parameter cases are covered, and in each case the only possible local minimum is the point $(x,y)$ mentionned above. Because the objective function and the contraints are all convex, the necessary conditions are also sufficient (since the Lagrangian is convex), so we have found the unique global minimum in each case.
Am I missing something? Thanks a lot for your help.