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.

notice that: $x - y = (x+z) - (y+z)$. So let $a = x+z$, and $b = y+z$, then: 
$ab = 1$, and minimize: $P(a,b) = \dfrac{1}{4(a-b)^2} + \dfrac{1}{a^2} + \dfrac{1}{b^2} = a^2 + \dfrac{1}{a^2} + \dfrac{1}{4(a^2 - 2 + b^2)} = a^2 + \dfrac{1}{a^2} + \dfrac{1}{4\left(a^2 + \dfrac{1}{a^2} - 2\right)}= t + \dfrac{1}{4t - 8}$, with $t = a^2 + \dfrac{1}{a^2} \geq 2$.
So now consider $P(a,b) = f(t) = t + \dfrac{1}{4t-8}$ on $2 \leq t < \infty$.
We have: $f'(t) = 1 -\dfrac{4}{(4t-8)^2} = 1 - \dfrac{1}{4t^2 - 16t + 16} = 0 \iff 4t^2 - 16t + 16 = 1 \iff (2t-4)^2 = 1 \iff 2t - 4 = 1 \iff t = \dfrac{5}{2} \iff a = \sqrt{2}$, and then $b = \dfrac{1}{\sqrt{2}}$.
So: $f_{min} = f(\frac{5}{2}) = 3 = P_{min}$
