Ta có: \(P=\frac{x^2+y^2}{x-y}=\frac{\left(x-y\right)^2+2xy}{x-y}=\left(x-y\right)+\frac{2xy}{x-y}\)
\(=x-y+\frac{16}{x-y}\ge2.4=8\)
Đặt \(t=x^2+y^2\) thì ta có :
\(P^2=\frac{\left(x^2+y^2\right)^2}{\left(x-y\right)^2}=\frac{t^2}{t-16}=\frac{1}{\frac{t-16}{t^2}}=\frac{1}{-\frac{16}{t^2}+\frac{1}{t}}=\frac{1}{-16\left(\frac{1}{t}-\frac{1}{32}\right)^2+\frac{1}{64}}\ge\frac{1}{\frac{1}{64}}=64\)
\(\Rightarrow P\ge8\). Đẳng thức xảy ra khi \(\hept{\begin{cases}x^2+y^2=32\\xy=8\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=2+2\sqrt{2}\\y=-2+2\sqrt{3}\end{cases}}\)
