Quadratic function, optimization, 2 variables

Question

I have the function

$m(x, y)=x^2+y^2+xy+3y+13, R(ρ)=\{x, y\}\ \in \mathbb{R}^2: -ρ\le\ x, y \le\ ρ$

I wanna show that if ρ>2 so that m has a local minimum within R(ρ) then this minimum is absolute and will always be inferior to any local minimum that is on the border of R(ρ). It is the last part of an exercise, it does not ask for a rigorous proof just a justification, but i am really lost.

Answer 1

$\frac {\partial m}{\partial x} = 2x + y = 0\\ \frac {\partial m}{\partial y} = x + 2y + 3 = 0\\ (x,y) = (1,-2)$

Is a critical point.

$m(x,y) = (x,y) \pmatrix {1&\frac 12\\ \frac 12 & 1} \pmatrix {x\\y} + 3y + 13$

That matrix is positive definite (has two positive eigenvalues) which means that $x,y$ moving away from the critical point only makes $m$ get bigger.

Or,

$m(x,y) = \frac 34 ((x-1)+(y+2))^2 + \frac 14((x-1)-(y+2))^2 +10$

Which clearly has a minimum at $(1,-2)$

Answer 2

$ x^2+y^2+xy+3y+13 = \left(x+\frac{y}{2}\right)^2 + \frac{3}{4} (y + 2)^2 + 10 \ge 10 $ with equality iff $y=-2, x=1$.