Đặt \(t=\dfrac{a^2+9}{a}\ge6\).
Ta có: \(P=7t+\dfrac{1}{t}=\left(7t+\dfrac{252}{t}\right)-\dfrac{251}{t}\ge_{AM-GM}2\sqrt{7.252}-\dfrac{251}{6}=84-\dfrac{251}{6}=\dfrac{253}{6}\).
Đẳng thức xảy ra khi và chỉ khi t = 6 \(\Leftrightarrow\dfrac{a^2+9}{a}=6\Leftrightarrow\left(a-3\right)^2=0\Leftrightarrow a=3\).
Vậy..
\(P=\dfrac{a^2+9}{36a}+\dfrac{a}{a^2+9}+\dfrac{251}{36}\left(\dfrac{a^2+9}{a}\right)\)
\(P\ge2\sqrt{\dfrac{\left(a^2+9\right).a}{36a\left(a^2+9\right)}}+\dfrac{251}{36}.\dfrac{2\sqrt{9a^2}}{a}=\dfrac{253}{6}\)
Dấu "=" xảy ra khi \(a=3\)