<s>[INST] Find Minimum value of $P=\frac{1}{1+2x}+\frac{1}{1+2y}+\frac{3-2xy}{5-x^2-y^2}$ Given: $x,y\in (-\sqrt2;\sqrt2)$ and $x^4+y^4+4=\dfrac{6}{xy}$
Find Minimum value Of $$P=\frac{1}{1+2x}+\frac{1}{1+2y}+\frac{3-2xy}{5-x^2-y^2}$$
Could someone help me ?
 [/INST] You could try this approach, although I think there is a simpler way.
Firstly, apply the change of variables 
$$ 
v=x^4+y^4, \ w = \frac{1}{xy}
$$
in order to simplify the domain $D$ to
$$
D = \{\ v(w) = 6w-4\ : w \in (-\infty,-1/2) \cup (1/ 2,\infty)\ \}.
$$
The minimal value is now given by invoking the chain rule
\begin{align}
0 =& \frac{dP}{dw}\\
  =& \frac{dP}{dv} \cdot \frac{dv}{dw} \\
  =& \Big(\frac{\partial P}{\partial x} \cdot \frac{dx}{dv} + \frac{\partial P}{\partial y} \cdot \frac{dy}{dv}\Big) \cdot \frac{dv}{dw} \\
  =& \Big( \frac{\partial P}{\partial x} \cdot \frac{dv}{dx}^{-1} + \frac{\partial P}{\partial y} \cdot \frac{dv}{dy}^{-1}\Big) \cdot \frac{dv}{dw},
\end{align}
taking into account that the solution lies in the domain $D$.
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