How to find the minimum of $xy$ from the equation $2x+y+6=xy$ , $x,y>0$

Question

I have tried doing factoring and completing the square and also AM-GM

$(2-y)(x-1)=-8$

$xy=2x+y+6≥\sqrt{12xy} $

Answer 1

AM-GM works perfectly fine with a small trick. $2x+y+6=xy \Rightarrow y(x-1)=2x+6 \Rightarrow y = \dfrac{2x+6}{x-1} \Rightarrow xy =\dfrac{2x^2+6x}{x-1}$.

As @Yves Daoust points out $\dfrac{2x^2+6x}{x-1} = 2x+8+ \dfrac{8}{x-1} = 2(x-1)+10+\dfrac{8}{x-1} $.

But by AM-GM, $ 2(x-1)+10+\dfrac{8}{x-1} \geq 10 + 2\left(\sqrt{2(x-1) \cdot \dfrac{8}{x-1} } \right) = 10+2\sqrt{16}=18$. This is indeed the minimum of $xy$.

Answer 2

Hint:

With $p:=xy$, after elimination of $y$ the equation can be written

$$p=\frac{2x^2+6x}{x-1},$$ with $x,p>0$.

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Note that

$$\frac{2x^2+6x}{x-1}=2x+8+\frac 8{x-1},$$ and finding the extrema is easy by differentiation.