Prove: $abc\geqslant 162$ [closed]

Given that $\frac{1}{a}+\frac{2}{b}+\frac{3}{c}=1$, prove that $abc\geqslant 162$.

I think that $abc=bc+2ac+3ab$, but I can't prove that $bc+2ac+3ab\geqslant162=3\cdot 6\cdot 9$

closed as off-topic by DeepSea, Mark Fantini, Shuchang, Travis, Claude LeiboviciDec 28 '14 at 10:04

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• Use AM-GM on $abc=bc+2ac+3ab \ge 3\sqrt[3]{6a^2b^2c^2}$ ! – r9m Dec 28 '14 at 8:49
• There are well known inequalities connecting various kinds of averages (weighted or not) – arithmetic, geometric, harmonic etc. Do you know these, and have you tried applying them? – Harald Hanche-Olsen Dec 28 '14 at 8:50

$$\huge \mathtt {GM\ge HM}$$ $$\mathtt {\left(a\cdot\frac b2\cdot\frac c3\right)^{\frac13}\ge \frac{3}{\displaystyle \frac{1}{a}+\frac{1}{\frac b2}+\frac{1}{\frac c3}}}\\\mathtt {abc\ge162}$$