<s>[INST] 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 ?
 [/INST] here is another way which is slightly different from Emil's.
$a=x+y,b=xy \implies 4b\le a^2 \iff 0<4a=4b(a-3) \le a^2(a-3) \iff 4\le a^2-3a \iff (a+1)(a-4)\ge 0 \iff a \ge 4, \\ P=a^2+\dfrac{3}{a}+1 $
note $f(t)=t^2+\dfrac{3}{t} =t(t-3)+3(t+\dfrac{1}{t})$ is mono increase function when $t>3$ because :
$t(t-3)$ is mono increasing function when $t>3$,
$t+\dfrac{1}{t}$ is mono increasing function when $t>1$
(indeed $f(t) \ge 3(t^2*(\dfrac{3}{2t})^2)^{\frac{1}{3}}=3(\dfrac{3}{2})^{\frac{2}{3}}, f(t) $is mono increasing function when $t \ge (\dfrac{3}{2})^{\frac{1}{3}}$)
$P_{min}=P(a=4)=4^2+\dfrac{3}{4}+1$
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