Find the minimum value of $P=\frac{1}{4(x-y)^2}+\frac{1}{(x+z)^2}+\frac{1}{(y+z)^2}$ Let $x,y,z$ be real numbers such that $x>y>0, z>0$ and $xy+(x+y)z+z^2=1$.
Find the minimum value of $$P=\frac{1}{4(x-y)^2}+\frac{1}{(x+z)^2}+\frac{1}{(y+z)^2}$$
I tried using some ways, but failed. Please give me an idea. Thank you.

From @DeepSea's above, suppose $b\equiv f(\,a\,),\,ab= 1\,\therefore\,{b}'= -\,\dfrac{b}{a},\,(\,a- 2\,b\,)(\,a- 2^{\,(\,\frac{1}{2}\,)}\,)\geqq 0$. 
$$a- b> 0 \tag{assume}$$
Let $$W(\,a\,)= \frac{1}{4(\,a- b\,)^{\,2}}+ \frac{1}{a^{\,2}}+ \frac{1}{b^{\,2}}- 3$$
$$\therefore\,{W}'(\,a\,)= -\,\frac{2}{a^{\,3}}+ \frac{{b}'- 1}{2(\,a- b\,)^{\,3}}- \frac{2\,{b}'}{b^{\,3}}$$
$$\therefore\,{W}'(\,a\,)= -\,\frac{2}{a^{\,3}}+ \frac{2}{ab^{\,2}}- \frac{-\,\dfrac{b}{a}- 1}{2(\,a- b\,)^{\,3}}$$
$$\therefore\,{W}'(\,a\,)= \frac{(\,a- 2\,b\,)(\,b- 2\,a\,)(\,a+ b\,)(\,2\,a^{\,2}- 3\,ab+ 2\,b^{\,2})}{2\,a^{\,3}b^{\,2}(\,a- b\,)^{\,2}}$$
$$\therefore\,(\,a- 2\,b\,){W}'(\,a\,)\geqq 0$$
$$\therefore\,(\,a- 2^{\,(\,\frac{1}{2}\,)}\,){W}'(\,a\,)\geqq 0$$
$$\therefore\,W(\,a\,)\geqq W(\,2^{\,(\,\frac{1}{2}\,)}\,)= 3$$
$$\because\,W(\,a\,)- W(\,2^{\,(\,\frac{1}{2}\,)}\,)= (\,a- 2^{\,(\,\frac{1}{2}\,)}\,){W}'(\,a\,) \tag{tangent equation}$$
