Question

Answer 1

-Đặt \(\left\{{}\begin{matrix}x=a-1>0\\y=b-1>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=x+1\\b=y+1\end{matrix}\right.\)

\(P=\dfrac{a^2}{a-1}+\dfrac{b^2}{b-1}=\dfrac{\left(x+1\right)^2}{x}+\dfrac{\left(y+1\right)^2}{y}\)

-Áp dụng BĐT AM-GM ta có:

\(\left\{{}\begin{matrix}\left(x+1\right)^2\ge4x\\\left(y+1\right)^2\ge4y\end{matrix}\right.\)

-Do đó: \(P=\dfrac{\left(x+1\right)^2}{x}+\dfrac{\left(y+1\right)^2}{y}\ge\dfrac{4x}{x}+\dfrac{4y}{y}=4+4=8\)

\(P=8\Leftrightarrow x=y=1\Leftrightarrow a=b=2\)

-Vậy \(P_{min}=8\)