Cách giải khác:
Dư đoán khi \(x=y=z=\frac{1}{\sqrt{3}}\) thì ta được \(P_{Min}=1\)
Thật vậy cần chứng minh \(Σ\frac{1}{4x^2-yz+2}\ge1\LeftrightarrowΣ\left(\frac{1}{4x^2-yz+2}-\frac{1}{3}\right)\ge0\)
\(\LeftrightarrowΣ\frac{1-4x^2+yz}{4x^2-yz+2}\ge0\LeftrightarrowΣ\frac{xy+xz+2yz-4x^2}{4x^2-yz+2}\ge0\)
\(\LeftrightarrowΣ\frac{\left(z-x\right)\left(2x+y\right)-\left(x-y\right)\left(2x+z\right)}{4x^2-yz+2}\ge0\)
\(\LeftrightarrowΣ\left(x-y\right)\left(\frac{2y+z}{4y^2-xz+2}-\frac{2x+z}{4x^2-yz+2}\right)\ge0\)
\(\LeftrightarrowΣ\left(x-y\right)^2\left(z^2+2xy+2\right)\left(z^2-xy+2\right)\ge0\)
Let \(\hept{\begin{cases}xy=a\\yz=b\\xz=c\end{cases}}\Rightarrow a+b+c=1\)
\(\Rightarrow x^2=\frac{ac}{b};y^2=\frac{ab}{c};z^2=\frac{bc}{a}\)
Hence \(S=Σ\frac{1}{\frac{4ac}{b}-b+2}=Σ\frac{1}{\frac{4ac}{b}-b+2\left(a+b+c\right)}\)
\(=Σ\frac{1}{\frac{4ac}{b}+2a+b+2c}=Σ\frac{b}{4ac+2ab+b^2+2bc}\)
\(=Σ\frac{b}{\left(2a+b\right)\left(2c+b\right)}\geΣ\frac{4b}{\left(2a+2b+2c\right)^2}=Σ\frac{b}{\left(a+b+c\right)^2}=1\)
dòng cuối sửa lại nhé
\(Σ\left(x-y\right)^2\left(z^2+2xy+2\right)\left(4z^2-xy+2\right)\ge0\)
