# Prove $\frac{1}{a^3} + \frac{1}{b^3} +\frac{1}{c^3} ≥ 3$

Prove inequality $$\frac{1}{a^3} + \frac{1}{b^3} +\frac{1}{c^3} ≥ 3$$ where $a+b+c=3abc$ and $a,b,c>0$

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Have you tried cross-multiplying (by $a^3b^3c^3$)? Say, first assumed that all of them are positive. –  Berci Nov 7 '12 at 14:28
Not a good idea Berci –  Xeing Nov 7 '12 at 14:44

If $a, b, c >0$ then $a+b+c=3abc \ \Rightarrow \ \cfrac 1{ab} + \cfrac 1{bc}+ \cfrac 1{ca} = 3$
See that $2\left(\cfrac 1{a^3} +\cfrac 1{b^3}+ \cfrac 1{c^3}\right) +3 =\left(\cfrac 1{a^3} +\cfrac 1{b^3}+ 1\right)+\left(\cfrac 1{b^3} +\cfrac 1{c^3}+ 1\right)+\left(\cfrac 1{c^3} +\cfrac 1{a^3}+ 1\right)$
Use $AM-GM$ inequality on each of them and you've got your proof.