Find the Min of P(x,y) Find the Minimum of the following function :
$$P(x,y) = \frac{(x-y)}{(x^4+y^4+6)}.$$
This is a math problem I found in an internet math competition but it is really complex to me !!!

Hint:  Armed with your guess, it is not hard to complete the square to get
$$\frac{x-y}{x^4+y^4+6}+\frac14= \frac{x^4+y^4+4x-4y+6}{4(x^4+y^4+6)}=\frac{(x^2-1)^2+(y^2-1)^2+2(x+1)^2+2(y-1)^2}{4(x^4+y^4+6)} \geqslant 0$$
with equality possible iff $x=-1, y=1$.
