<s>[INST] $\min\left(\sum_{\text{sym}}\frac{(2x^2+y)(4x + y^2)}{\underset{\ne0}{(2x + y - 2)^2}} - 3(\underset{\gt0}{x}+\underset{\gt0}{y})\right)=?$ 
Given positives $x$ and $y$ such that $2x + y \ne 2$ and $x + 2y \ne 2$, calculate the minimum value of $$\frac{(2x^2 + y)(4x + y^2)}{(2x + y - 2)^2} + \frac{(2y^2 + x)(4y + x^2)}{(2y + x - 2)^2} - 3(x + y)$$

We have that $$\frac{(2x^2 + y)(4x + y^2)}{(2x + y - 2)^2} + \frac{(2y^2 + x)(4y + x^2)}{(2y + x - 2)^2} - 3(x + y)$$
$$ = \sum_{\text{sym}}\left[\frac{(2x^2 + y)(4x + y^2)}{(2x + y - 2)^2} - (2x + y)\right] = 2 \cdot \sum_{\text{sym}}\frac{(xy - 2x - y)(xy - 4x - 2y + 2)}{(2x + y - 2)^2}$$
$$ = 2 \cdot \sum_{\text{sym}}\left[\left(\frac{xy - 2x - y}{2x + y - 2}\right)^2 - \frac{xy - 2x - y}{2x + y - 2}\right] \ge 2 \cdot \left[2 \cdot \left(-\frac{1}{4}\right)\right] = -1$$
The equal sign occurs when $\dfrac{xy - 2x - y}{2x + y - 2} = \dfrac{xy - 2y - x}{2y + x - 2} = \dfrac{1}{2}$
$$\iff \dfrac{xy - 2}{2x + y - 2} = \dfrac{xy - 2}{2y + x - 2} = \dfrac{3}{2}$$
$$\implies 2x + y - 2 = 2y + x - 2 \iff x = y$$
Furthermore, $\dfrac{xy - 2}{2x + y - 2} = \dfrac{xy - 2}{2y + x - 2} = \dfrac{3}{2}$
$$\iff 2(xy - 2) = 3(2x + y - 2) = 3(2y + x - 2)$$
$$\implies 2(x^2 - 2) - 3(3x - 2) = 2(y^2 - 2) - 3(3y - 2) = 0$$
$$\implies 2x^2 - 9x + 2 = 2y^2 - 9y + 2 = 0 \implies x = y = \frac{9 + \pm \sqrt{65}}{4}$$
I would like to know whether the above solution is correct and if not, whether there are any other solutions.
Thanks for your attention.
 [/INST] Solution by Nguyen Van Quy.
First, we have: If $a,b>0,\ a+b\neq 1$ then $$\dfrac{2(a^2+b)(a+b^2)}{(a+b-1)^2} \geq \dfrac{4a+4b-1}{2}.$$ Indeed, inequality equivalent to
$$4(a^3+b^3+a^2b^2+ab)\geq (4a+4b-1)(a+b-1)^2,$$ or
$$4\big[(a+b)^3-3ab(a+b)+a^2b^2+ab\big]\geq
4(a+b)^3-9(a+b)^2+6(a+b)-1,$$ or $$9(a+b)^2-6(a+b)(2ab+1)+(2ab+1)^2\geq
0,$$ or $$(3a+3b-2ab-1)^2\geq 0.$$ Done.
Now replace $(a,b)$ to $\left(x,\dfrac{y}{2}\right)$ and $\left(\dfrac{x}{2}, y\right),$ we get
$$\dfrac{(2x^2+y)(4x+y^2)}{(2x+y-2)^2} \geq 2x+y-\dfrac{1}{2},$$
$$\dfrac{(2y^2+x)(4y+x^2)}{(x+2y-2)^2} \geq 2y+x-\frac{1}{2}.$$
Therefore
$$\frac{(2x^2 + y)(4x + y^2)}{(2x + y - 2)^2} +\frac{(2y^2 + x)(4y + x^2)}{(2y + x - 2)^2} \geqslant 3(x + y) -1,$$
or
$$\frac{(2x^2 + y)(4x + y^2)}{(2x + y - 2)^2} +\frac{(2y^2 + x)(4y + x^2)}{(2y + x - 2)^2} - 3(x + y) \geqslant -1.$$
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