Question

Answer 1

Giả thiết tương đương: \(\left(x+y\right)^2-2xy-\left(x+y\right)=xy\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)

\(\Rightarrow u^2-u=3v\) 

Khi đó: \(P=\left(x+y\right)^2-2xy-\dfrac{8}{3}\sqrt{x+y}=u^2-\dfrac{2}{3}\left(u^2-u\right)-\dfrac{8}{3}\sqrt{u}\)

\(P=\dfrac{1}{3}u^2+\dfrac{2}{3}u-\dfrac{8}{3}\sqrt{u}=\dfrac{1}{3}t^4+\dfrac{2}{3}t^2-\dfrac{8}{3}t\) với \(t=\sqrt{u}\ge0\)

\(f'\left(t\right)=\dfrac{4}{3}t^3+\dfrac{4}{3}t-\dfrac{8}{3}=0\Rightarrow t=1\)

BBT \(\Rightarrow f\left(t\right)_{min}=f\left(1\right)=-\dfrac{5}{3}\)