Lời giải:
Áp dụng BĐT Cauchy:
\(\left\{\begin{matrix} 9b+a\geq 6\sqrt{ab}\\ 8a+2c\geq 8\sqrt{ac}\end{matrix}\right.\Rightarrow 6\sqrt{ab}+8\sqrt{ac}+7c\leq 9(a+b+c)\)
Do đó \(P\geq \frac{1}{9(a+b+c)}+2\sqrt{a+b+c}\)
Tiếp tục áp dụng BĐT Cauchy:
\(\frac{1}{9(a+b+c)}+\frac{\sqrt{a+b+c}}{243}+\frac{\sqrt{a+b+c}}{243}\geq 3\sqrt[3]{\frac{1}{9.243.243}}=\frac{1}{27}\)
Và \(\frac{484\sqrt{a+b+c}}{243}\geq \frac{484}{81}\) do \(a+b+c\geq 9\)
Cộng theo vế suy ra \(P\geq \frac{1}{9(a+b+c)}+2\sqrt{a+b+c}\geq \frac{1}{27}+\frac{484}{81}=\frac{487}{81}\)
Vậy \(P_{\min}=\frac{487}{81}\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} a=9b\\ 4a=c\\ a+b+c=9\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a=\frac{81}{46}\\ b=\frac{9}{46}\\ c=\frac{162}{23}\end{matrix}\right.\)
