Find the largest value of $P=\frac{3x+2y+1}{x+y+6}$ I know there has been a similar question here, but my question is a little bit different
The Problem:
Given that $3(x+y)=x^2+y^2+xy+2$,
then find the maximum value of
\begin{align}
P=\frac{3x+2y+1}{x+y+6}
\end{align}
What I was thinking about is that I'm trying to transform P into a first-degree equation on $x$, which is totally possible using the condition in the problem, and then apply the basic Cauchy inequalities.
However, I haven't been able to figure it out. The other possibility is to apply differentiation like in the problem I mentioned at the beginning of the question. The problem is that I took this problem from Olympiad training, which often does not use differentiation (so there should be a way that doesn't use the method)
So any help is appreciated
 A: I would apply the following technique.  It may not be the best, but I can definitely say it's working:
$\underline {\text{Case - 1:}}$ $\thinspace \thinspace \thinspace \thinspace \thinspace P=2.$
If $P=2$, we get $x=11$. This implies,
$$(y + 4)^2 + 74 = 0$$ which gives a contradiction.
$\underline{\text{Case - 2:}}$ $\thinspace \thinspace \thinspace \thinspace \thinspace P≠2$.
We have,
$$\begin{align}&2y-Py+3x-Px-6P+1=0\\
\implies &y=\frac{Px-3x+6P-1}{2-P},\thinspace P≠2\end{align}$$
Then, substituting $y=\frac{Px-3x+6P-1}{2-P}$ for $y$ in the
polynomial equation $x^2+y^2+xy-3x-3y+2=0$, we get a quadratic equation with respect to $x:$
$$\begin{align}&x^2(P^2-5P+7)+x(6P^2-28P+10)+(56P^2-59P+15)=0\\
\implies &\Delta_x=- (47 P^4 - 255 P^3 + 476 P^2 - 348 P + 80)≥0\\
\implies &\Delta_x=-(P - 2)^2 (P - 1) (47 P - 20)≥0,\thinspace P≠2\\
\implies &\frac {20}{47}≤P≤1\\
\implies &\min\left\{P\right\}=\frac{20}{47}\\
\implies &\max\left\{P\right\}=1.\end{align}$$
A: $3(x+y)=x^{2}+y^{2}+xy+2$ must be a conic section because it can be written in the form $Ax^2+Bxy+Cy^2+Dx+Ey+F = 0$. It also is symmetric across the line $y=x$ as it is invariant after swapping $x$ and $y$.
Now if the maximum value is $P$, then $3x+2y+1 = P(x+y+6)$, equation $1$. This can be seen below:

If we can rotate this ellipse by $45º$, we can write it in parametric form in terms of an angle $\theta$. Using the rotation matrix and a scaling factor, we need the transformation $x \to x - y, y \to x + y$ as $\cos \pi/4 = \sin \pi/4$:
$$3(x-y + x+y) = (x-y)^2 + (x+y)^2 + (x-y)(x+y) + 2$$
$$6x = 2(x^2 + y^2) + (x^2-y^2) + 2$$
$$6x = 3x^2 + y^2 + 2 \tag{2}$$
where from $(1)$, we now have:
$$3(x-y) + 2(x+y) + 1 = P \left( (x-y) + (x+y) + 6 \right)$$
$$5x-y+1 = P(2x+6)$$
$$y = (5-2P)x-6P+1 \tag{3}$$
Just to check, the lines are still tangent to the ellipse after transformation:

and now we can substitute $(3)$ into $(2)$:
$$6x = 3x^2 + ((5-2P)x-6P+1)^2 + 2$$
$$6x = 3x^2 + (25-20P+4P^2)x^2 + 2(5-2P)(-6P+1)x + 36P^2-12P+1+2$$
$$(28-20P+4P^2)x^2 + (24P^2-64P+4)x + 36P^2-12P+3 = 0$$
$$\Delta = 0: (24P^2-64P+4)^2 - 4(28-20P+4P^2)(36P^2-12P+3) = 0$$
$$4(6P^2-16P+1)^2 - 48(7-5P+P^2)(P^2-4P+1) = 0$$
and miraculously, this is a quadratic $-752P^2 + 1072P - 320 = 0 \implies (47-20P)(P-1) = 0$, so the maximum value is $P = 1$.
A: For $x=2$ and $y=1$ we obtain a value $1$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$\frac{3x+2y+1}{x+y+6}\leq1$$ and since by the condition $x+y+6>0,$ we need to prove that:
$$2x+y\leq5.$$
Indeed, by the condition again and by C-S we obtain:
$$14=(3-x)^2+(x+y)^2+(3-y)^2=$$
$$=\frac{1}{14}(1^2+3^2+2^2)\left((3-x)^2+(x+y)^2+(3-y)^2\right)\geq$$
$$\geq\frac{1}{14}(3-x+3x+3y+6-2y)^2=\frac{1}{14}(2x+y+9)^2,$$ which gives
$$2x+y+9\leq14$$ or
$$2x+y\leq5$$ and we are done.
