<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] Let $x=y=-1$.
Thus, we obtain the value $16$.
We'll prove that it's a minimal value.
Indeed,
$$(5+x)(5+y)-16=xy+5(x+y)+9=3x^2+4xy+3y^2+5(x+y)\sqrt{\frac{x^2+y^2+xy}{3}}\geq$$
$$\geq3x^2+4xy+3y^2-5\sqrt{\frac{(x+y)^2(x^2+y^2+xy)}{3}}=$$
$$=\frac{3(3x^2+4xy+3y^2)^2-25(x+y)^2(x^2+y^2+xy)}{\sqrt3(\sqrt{3}(x^2+y^2+xy)+5\sqrt{(x+y)^2(x^2+y^2+xy)})}=$$
$$=\frac{(x-y)^2(2x^2+xy+2y^2)}{\sqrt3(\sqrt{3}(x^2+y^2+xy)+5\sqrt{(x+y)^2(x^2+y^2+xy)})}\geq0$$ and we are done!
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