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

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 $x=-1, y=-1$.
