Ta có: \(\left(x+z\right)\left(y+z\right)=1\)
\(\Rightarrow\left(x+z\right)^2\left(y+z\right)^2=1\)
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{\left(y+z\right)^2}+\dfrac{1}{\left(z+x\right)^2}=\dfrac{1}{\left(x-y\right)^2}+\dfrac{\left(x+z\right)^2\left(y+z\right)^2}{\left(y+z\right)^2}+\dfrac{\left(x+z\right)^2\left(y+z\right)^2}{\left(z+x\right)^2}\)
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x+z\right)^2+\left(y+z\right)^2\)
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x+z\right)^2-2\left(x+z\right)\left(y+z\right)+\left(y+z\right)^2+2\) (Vì: (x+z)(y+z)=1 =>2(x+z)(y+z)=2 )
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x+z-y-z\right)^2+2\)
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x-y\right)^2+2\)
Áp dụng bất đẳng thức Cauchy, ta có :
\(\dfrac{1}{\left(x-y\right)^2}+\left(x-y\right)^2\ge2\sqrt{\dfrac{1}{\left(x-y\right)^2}\cdot\left(x-y\right)^2}=2\cdot1=2\)
\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x-y\right)^2+2\ge2+2=4\)
Vậy \(MinP=4\) khi \(x-y=1\); \(y+z=\dfrac{\sqrt{5}-1}{2}\); \(x+z=\dfrac{2}{\sqrt{5}-1}\)
