Another system of equations

Let us deploy Lagrange Multipliers here. If $f(x,y) = x^2 + xy + y^2$ and $g(x,y) = 2x^2 + xy - y^2 = 1$ , then we have $\nabla f = \lambda \cdot \nabla g$ , or:

$2x + y = \lambda (4x+y)$ ;

$x+2y = \lambda (x-2y)$

or $\frac{2x+y}{4x+y} = \frac{x+2y}{x-2y} \Rightarrow 2x^2 - 3xy -2y^2 = 4x^2 + 9xy + 2y^2 \Rightarrow x^2 + 6xy + 2y^2 = 0 \Rightarrow x^2 + 6xy + 9y^2 = 7y^2 \Rightarrow (x+3y)^2 = 7y^2 \Rightarrow x = (-3 \pm \sqrt{7})y.$

Taking $g((-3+\sqrt{7})y,y) = 2(-3+\sqrt{7})^2y^2 + (-3+\sqrt{7})y^2 - y^2 = 1\ \Rightarrow y^2 = \frac{1}{\sqrt{28-11\sqrt{7}}} \Rightarrow y = \pm\frac{i}{\sqrt{28-11\sqrt{7}}}$ (which are imaginary).

Taking $g((-3-\sqrt{7})y,y) = 2(-3-\sqrt{7})^2y^2 + (-3-\sqrt{7})y^2 - y^2 = 1\ \Rightarrow y^2 = \frac{1}{\sqrt{28+11\sqrt{7}}} \Rightarrow y = \pm\frac{1}{\sqrt{28+11\sqrt{7}}}$ (which are real).

The Hessian matrix of $f$ computes to:

$F(x,y) = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$

which has eigenvalues $\det(F-I\alpha) = 0 \Rightarrow (2-\alpha)^2 -1 = 0 \Rightarrow \alpha = 2 \pm 1 = 1,3$ that are real, distinct, & positive $\Rightarrow F$ is positive-definite for all $x,y \in \mathbb{R}.$

Hence, we have the real pairs $\large (x,y) = (\frac{3+\sqrt{7}}{\sqrt{28+11\sqrt{7}}}, -\frac{1}{\sqrt{28+11\sqrt{7}}}); (-\frac{3+\sqrt{7}}{\sqrt{28+11\sqrt{7}}}, \frac{1}{\sqrt{28+11\sqrt{7}}})$ , and substitution of either pair into $f$ yields the minimum value:

$\large m \ge (\frac{3+\sqrt{7}}{\sqrt{28+11\sqrt{7}}})^2 + (\frac{3+\sqrt{7}}{\sqrt{28+11\sqrt{7}}})(-\frac{1}{\sqrt{28+11\sqrt{7}}}) + (-\frac{1}{\sqrt{28+11\sqrt{7}}})^2;$

or $\large m \ge \frac{16 + 6\sqrt{7} - 3 - \sqrt{7} + 1}{28+11\sqrt{7}} = \frac{14 + 5\sqrt{7}}{28+11\sqrt{7}} \cdot \frac{28-11\sqrt{7}}{28-11\sqrt{7}} = \boxed{\frac{-1+2\sqrt{7}}{9}}.$

or $a = -1, b =2, c=7, d = 9 \Rightarrow a+b+c+d = -1 + 2+7+9 = \boxed{17}.$