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
 A: 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 $16$.
A: 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 $x=-1, y=-1$.
A: 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.
A: 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!
A: Using Rotation of axes
if $t$ is the rotation of axes, $\cot2t=0\iff\cos2t=0$
Set $2t=\dfrac\pi2$
$\sqrt2u=x+y,\sqrt2v=y-x$
$$3=\dfrac{(u-v)^2+(u+v)^2}2=u^2+v^2$$
WLOG $u=\sqrt3\cos t, v=\sqrt3\sin t$
$$(5+x)(5+y)=25+\sqrt2u+\dfrac{2(u^2-v^2)}4=\dfrac{50+2\sqrt2u+u^2-v^2}2$$
$2\sqrt2u+u^2-v^2=2\sqrt2(\sqrt3\cos t)+3(\cos^2t-\sin^2t)$
$=6\cos^2t+2\sqrt6\cos t-3=6\left(\cos t+\dfrac1{\sqrt6}\right)^2-3-6\left(\dfrac1{\sqrt6}\right)^2$
Now $-1\le\cos t\le1\iff -1+\dfrac1{\sqrt6}\le\cos t+\dfrac1{\sqrt6}\le1+\dfrac1{\sqrt6}$
$\implies\left(\cos t+\dfrac1{\sqrt6}\right)^2\le\left(1+\dfrac1{\sqrt6}\right)^2$
