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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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    $\begingroup$ Use AM-GM on $abc=bc+2ac+3ab \ge 3\sqrt[3]{6a^2b^2c^2}$ ! $\endgroup$
    – r9m
    Dec 28, 2014 at 8:49
  • $\begingroup$ 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? $\endgroup$ Dec 28, 2014 at 8:50

1 Answer 1

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

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