Áp dụng BĐT Cauchy Schwarz ta có:
$P\ge\dfrac{(\sqrt a+\sqrt b+\sqrt c)^2}{2(\sqrt a+\sqrt b+\sqrt c)-9}$
Đặt $t=\sqrt a+\sqrt b+\sqrt c$.
Ta có: $t^2+81\ge18t \Leftrightarrow t^2\ge9(2t-9) \Leftrightarrow \dfrac{t^2}{2t-9}\ge9$
Từ đó suy ra $P\ge9$
$\min P=9 \Leftrightarrow \left\{\begin{array}{l}a=b=c\\t=9\end{array}\right. \Leftrightarrow a=b=c=9$