Minimum Value

If $(x+y)^2+(x-y)^2 = 2(x^2+y^2)$ , then let us solve this NLP according to:

MIN

$2(x^2 + y^2)$

subject to:

$y \le -x^2 - 2x;$

$y \ge -\frac{2}{3}x + \frac{1}{3}$ .

The two curves intersect at the points $-x^2-2x = -\frac{2}{3}x + \frac{1}{3} \Rightarrow x^2 + \frac{4}{3}x + \frac{1}{3} = (x+1)(x+1/3) = 0 \Rightarrow (x,y) = (-1,1); (-1/3, 5/9),$ which are the critical vertices of the feasible region below:

Plugging in the above two critical vertices into our objective function produces:

$(-1,1) \Rightarrow 2(1+1) = 4$ ;

$(-1/3, 5/9) \Rightarrow 2(\frac{1}{9}+\frac{25}{81}) = \frac{68}{81}$

hence, the minimum value is $\boxed{\frac{68}{81}}.$