Question

Cho \(x,y,z\in R\) và \(\left(x-y\right)\left(x-z\right)=1\) với \(y\ne z\)

CM: \(S=\frac{1}{\left(x-y\right)^2}+\frac{1}{\left(y-z\right)^2}+\frac{1}{\left(z-x\right)^2}\ge4\)

Answer 1

Đặt \(\left\{{}\begin{matrix}x-y=a\\x-z=b\end{matrix}\right.\) \(\Rightarrow ab=1\)

\(S=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a-b\right)^2}=\frac{a^2+b^2}{a^2b^2}+\frac{1}{\left(a-b\right)^2}=a^2+b^2+\frac{1}{\left(a-b\right)^2}\)

\(S=a^2+b^2-2ab+\frac{1}{\left(a-b\right)^2}+2=\left(a-b\right)^2+\frac{1}{\left(a-b\right)^2}+2\)

\(S\ge2\sqrt{\frac{\left(a-b\right)^2}{\left(a-b\right)^2}}+2=4\) (đpcm)