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

Hint: From the equation $$y^2+xy+x^2-3=0$$ we get 
$$y_{1,2}=-\frac{x}{2}\pm\sqrt{3-\frac{3}{4}x^2}$$ so you will obtain the two functions 
 $$f_1(x)=(5+x)\left(5-\frac{x}{2}+\sqrt{3-\frac{3}{4}x^2}\right)$$
or
$$f_2(x)=(5+x)\left(5-\frac{x}{2}-\sqrt{3-\frac{3}{4}x^2}\right)$$
to consider.
Doing this we get $$(5+x)(5+y)\geq 16$$, the equal sign holds if $$x=y=-1$$ as Mr. Rozenberg stated.
