Xin ảnh:v
\(P=\frac{x^2}{y-1}+\frac{y^2}{x-1}\ge2\sqrt{\frac{x^2y^2}{\left(y-1\right)\left(x-1\right)}}=2\sqrt{\frac{x^2}{x-1}.\frac{y^2}{y-1}}\)
Ta có: \(\frac{x^2}{x-1}=\frac{x^2-4x+4+4x-4}{x-1}=\frac{\left(x-2\right)^2}{x-1}+4\ge4\) (đúng với mọi x>1)
Tương tự vs y ta có: \(2\sqrt{\frac{x^2}{x-1}.\frac{y^2}{y-1}}\ge2\sqrt{4.4}=8\)
\("="\Leftrightarrow x=y=2\)
\(P\ge\frac{\left(x+y\right)^2}{x+y-2}=x+y+2+\frac{4}{x+y-2}=x+y-2+\frac{4}{x+y-2}+4\)
\(P\ge2\sqrt{\left(x+y-2\right).\frac{4}{\left(x+y-2\right)}}+4=8\)
\(\Rightarrow P_{min}=8\) khi \(\left\{{}\begin{matrix}x=y\\x+y-2=2\end{matrix}\right.\) \(\Rightarrow x=y=2\)
