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 !!!
 A: 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 $x=-1, y=1$.
A: It is not difficult to see that the minimum is located at
$x<0$ and $y>0$, so by the symmetry, it suffices to consider the problem
$$\max P(x,y)\quad\mbox{s.t.}\quad x,y\geq 0.$$
Now we use the polar coordinate as below.
Let $x=r\cos\theta$ and $y=r\sin\theta$, where $r\geq 0$ and 
$0\leq\theta\leq\frac{\pi}{2}$. Then
\begin{align*}
P(x,y)\equiv\bar{P}(r,\theta)&=\frac{r(\cos\theta+\sin\theta)}{r^4(\cos^4\theta+\sin^4\theta)+6}\\
&=\frac{\sqrt{2}r\cos(\theta-\frac{\pi}{4})}{r^4[(\cos^2\theta+\sin^2\theta)^2-2\sin^2\theta\cos^2\theta]+6}\\
&=\frac{\sqrt{2}r\cos(\theta-\frac{\pi}{4})}{r^4[1-\frac{1}{2}\sin^2(2\theta)]+6}\\
&=\frac{\sqrt{2}r\cos(\theta-\frac{\pi}{4})}{\frac{r^4}{4}[3+\cos(4\theta)]+6}.
\end{align*}
We see that when fixing $r$, $\bar{P}$ reaches to maximum possibly when $\theta=\frac{\pi}{4}$. That is,
$$\bar{P}(r,\theta)\leq\frac{\sqrt{2}r}{\frac{r^4}{2}+6}=\frac{2\sqrt{2}r}{r^4+12}\equiv Q(r).$$
Finally, we calculate 
$$Q'(r)=\frac{2\sqrt{2}(r^4+12)-8\sqrt{2}r^4}{(r^4+12)^2}=0
\quad\Longrightarrow\quad r=\sqrt{2}.$$
By checking $Q''(\sqrt{2})<0$, we conclude that
$P(1,1)=\bar{P}(\sqrt{2},\frac{\pi}{4})$ is the maximum. Hence $P(-1,1)=-\frac{1}{4}$ is the minimum.
