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 $bc+2ac+3ab\geqslant162=3\cdot 6\cdot 9$
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Sign up to join this communityGiven 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 $bc+2ac+3ab\geqslant162=3\cdot 6\cdot 9$
$$\huge \mathtt {GM\ge HM}$$ $$\mathtt {\left(a\cdot\frac b2\cdot\frac c3\right)^{\frac13}\ge \frac{3}{\displaystyle \frac{1}{a}+\frac{1}{\frac b2}+\frac{1}{\frac c3}}}\\\mathtt {abc\ge162}$$