inequality

**Ortega** · March 7th 2009, 02:41 PM

if abc=1 then demonstrate:
$\frac{a}{b} + \frac{b}{c} + \frac{1}{a} \ge a + b + 1$

**horan** · March 7th 2009, 08:12 PM

This should help:
$<br /> \frac{ac}{ac} \frac{a}{b} + \frac{ab}{ab} \frac{b}{c} + \frac{bc}{bc}\frac{1}{a} = <br /> \frac{a^2c}{abc} + \frac{b^2a}{abc} + \frac{bc}{abc} = \frac{a^2c}{1} + \frac{b^2a}{1} + \frac{bc}{1}<br /> = a^2 c + b^2 a + bc<br />$

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