minimum value of of $(5+x)(5+y)$

The feasible set is a (compact) ellipse $E$ in the $(x,y)$-plane. We have to consider the Lagrangian $$\Phi:=(x+5)(y+5)-\lambda(x^2+xy+y^2-3)\ .$$ The equations $$\Phi_x=y+5-\lambda(2x+y)=0,\qquad\Phi_y=x+5-\lambda(x+2y)=0\ ,$$ or $$\left\{\eqalign{2\lambda x+(\lambda-1)y&=5 \cr (\lambda-1)x+2\lambda y&=5\cr}\right.$$ are symmetric in $x$ and $y$, hence the solution is given by $$x=y={5\over3\lambda-1}\ .\tag{1}$$ Plugging this into the equation $x^2+xy+y^2=3$ of $E$ gives $(3\lambda-1)^2=25$, or $3\lambda-1=\pm5$. From $(1)$ we then get two conditionally stationary points of $f$ on $E$, namely $(1,1)$ and $(-1,-1)$. From $f(1,1)=36$ and $f(-1,-1)=16$ it then follows that the minimum of $f$ on $E$ is $16$.

The symmetry of the problem suggests to substitute $x = u+v$, $y = u-v$. Then the given constraint is $$ 3 = x^2 + xy + y^2 = 3u^2 + v^2 $$ and in particular $-1 \le u \le 1$.

We look for the minimal value of $$ (5+x)(5+y) = (5+u)^2 - v^2 = (5+u)^2 - (3-3u^2) \\ = 4u^2 + 10 u + 22 \, . $$ This expression is increasing for $u \in [-1, 1]$, and therefore $$ \ge 4 (-1)^2 + 10(-1) + 22 = 16 \, . $$ Equality holds for $u=-1, v=0$, corresponding to $x=-1, y=-1$.