Prove inequality $\frac{1}{a^3} + \frac{1}{b^3} +\frac{1}{c^3} ≥ 3$ where $a+b+c=3abc$ and $a,b,c>0$
Prove $\frac{1}{a^3} + \frac{1}{b^3} +\frac{1}{c^3} ≥ 3$
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inequality
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0Not a good idea Berci – 2012-11-07
1 Answers
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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.
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0Good and Fast answer – 2012-11-07