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 ?
 A: It is sure that, without using Lagrange multipliers, the problem is quite hard.
But let us try using brute force. The constraint $x^2y-x+xy^2-y-3xy=0$ gives a quadratic equation in $y$; so we can express $y$ as a function of $x$ $$y_{\pm}=\frac{-x^2+3 x+1\pm\sqrt{x^4-6 x^3+11 x^2+6 x+1}}{2 x}$$ Replace $y_+$ in $P$ (which is now an ugly function of $x$) and plot it (even better, compute $P'_x$ and plot it to find the root). You will effectively find $x=2$ to which corresponds $y=2$ and $P=\frac{71}{4}$.
Doing the same with $y_-$ shows a minimum for $x\approx 1.54426$ corresponding to $y \approx-0.399545$ which has to be discarded.
A: $x^2y-x+xy^2-y-3xy=0 , x>0, y>0$
$xy(x+y)-(x+y)-3xy=0$
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 $a\ge 4$)!
A: 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 $t \ge (\dfrac{3}{2})^{\frac{1}{3}}$)
$P_{min}=P(a=4)=4^2+\dfrac{3}{4}+1$
