<s>[INST] minimum value of of $(5+x)(5+y)$ 
If $x^2+xy+y^2=3$ and $x,y\in \mathbb{R}.$
Then find the minimum value of $(5+x)(5+y)$

What I try
$$(5+x)(5+y)=25+5(x+y)+xy$$
$x^2+xy+y^2=3\Rightarrow (x+y)^2-3=xy$
I am finding  $f(x,y)=22+5(x+y)+(x+y)^2$
How do I solve it? Help me please
 [/INST] 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 $16$.
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