Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(a+b+c)\geq (1+1+1)^2\)
\(\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c}=9\)
Vậy $P_{\min}=9$ khi $a=b=c=\frac{1}{3}$
Hoặc cách khác:
Áp dụng BĐT Cô-si:
\(\frac{1}{a}+9a\geq 2\sqrt{\frac{1}{a}.9a}=6\)
\(\frac{1}{b}+9b\geq 6\)
\(\frac{1}{c}+9c\geq 6\)
Cộng theo vế: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+9(a+b+c)\geq 18\)
\(\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+9\geq 18\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 9\)
Vậy $P_{\min}=9$
