Find Minimum Value of: $P=x^2+y^2+2xy+2-\frac{1}{xy}$ Given: $x,y>0$ and $x^2y-x+xy^2-y-3xy=0$
Find Minimum Value of:
$P=x^2+y^2+2xy+2-\frac{1}{xy}$
I found $\min P =\frac{71}{4}$ at $x=y=2$ but I cant prove that
Could some one help me ?

$x^2y-x+xy^2-y-3xy=0 , x>0, y>0$
$xy(x+y)-(x+y)-3xy=0$
Let $\begin{array}{|l} {x+y=a} \\ {xy=b} \end{array} \Rightarrow a>0, b>0$
Then $ab-a-3b=0$
$b(a-3)=a \Rightarrow a-3>0 \Rightarrow a>3$
$b=\frac{a}{a-3 }$
x and y are roots of $t^2-at+b=0$
$D=a^2-4b\ge 0$
$a^2-\frac{4a}{a-3 }\ge 0$ |.$(a-3)>0$
$a^3-3a^2-4a\ge 0$
$a(a^2-3a-4)\ge 0$ |:$a>0$
$a^2-3a-4\ge 0$
$(a-4)(a+1)\ge 0$ |:$(a+1)>0$
$a-4\ge 0 \Rightarrow a\ge 4$
$P=x^2+y^2+2xy+2-\frac{1}{xy }=(x+y)^2-\frac{1}{xy }+2=a^2-\frac{1}{b }+2=$
$=a^2-\frac{a-3}{a }+2= a^2+\frac{3}{a }+1$
$a^2+\frac{3}{a }+1\ge \frac{71}{4 } \Leftrightarrow 4a^3-67a+12\ge 0 \Leftrightarrow (a-4)(4a^2+16a-3)\ge 0$
DONE ($4a^2+16a-3\ge 0$ if $a\ge 4$)!
