Question

Cho x,y,z là các số thực thỏa mãn (x-y)(x-z)=1; y ≠ z.

Chứng minh rằng: \(\frac{1}{\left(x-y\right)^2}+\frac{1}{\left(y-z\right)^2}+\frac{1}{\left(z-x\right)^2}\ge4\)

Answer 1

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

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

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

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

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\left(x-y\right)\left(x-z\right)=1\\\left(y-z\right)^2=1\end{matrix}\right.\)