Thay a = b+1
\(P=\frac{\left(b+1\right)^2+b^2}{b}=\frac{2b^2+2b+1}{b}=2b+2+\frac{1}{b}\ge2+2\sqrt{2b.\frac{1}{b}}=2+2\sqrt{2}\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}a=b+1\\2b=\frac{1}{b}\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}a=\frac{2+\sqrt{2}}{2}\\b=\frac{\sqrt{2}}{2}\end{cases}}\)
Vậy \(P_{min}=2+2\sqrt{2}\)