$\frac 1a+\frac 2b+\frac 3c \overset{AM-GM}{\ge} \frac 6{\sqrt[6]{ab^2c^3}}$$\Leftrightarrow \sqrt[6]{ab^2c^3} \ge \frac{6}{\dfrac 1a+\dfrac 2b+\dfrac 3c}$ (1)
Lại có
$\frac 1a+\frac 2b+\frac 3c = \left( \frac 1a-1 \right) + \left( \frac 2b -2\right)+\left( \frac 3c-3 \right)+6$
$=\frac{1-a}{a}+\frac{2(1-b)}{b}+\frac{3(1-c)}{c}+6 \le \frac{1-a}{a}+\frac{2(1-b)}{a}+\frac{3(1-c)}{a}+6$
$=\frac{6-(a+2b+3c)}{a}+6 \le \frac{6-2(a+b+c)}{a}+6$
$=\frac{6-2\sqrt{(a+b+c)(\frac 1a+\frac 1b+\frac 1c)}}{a}+6 \le \frac{6-2.\sqrt 9}{a}+6=6$ (2)
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Từ (1) & (2) $\Rightarrow \sqrt[6]{ab^2c^3} \ge 1\Rightarrow P \ge 1$
Vậy $\min P=1\Leftrightarrow a=b=c=1$