# prove that $\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\geq a+b+c$ [duplicate]

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If $a,b,c>0$, Then prove that $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\geq a+b+c$$

$\bf{My\; Try::}$ Using Cauchy- Schwarz Inequality

$$\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\geq \frac{a^2+b^2+c^2}{3abc}$$

Now How can i solve after that , Help required, Thanks

## marked as duplicate by Arnaud D., mrp, Davide Giraudo, Stefan4024, John BOct 9 '16 at 22:53

$$a^4+b^4+c^4\geq a^2bc+b^2ac+c^2 ab$$ holds by the rearrangement inequality or Muirhead's inequality.
$$(b+c+a)(c+a+b) \Big(\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\Big) \geqslant (a+b+c)^3$$
From where you get: $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab} \geqslant a+b+c$$