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- #1

- Feb 14, 2012

- 3,802

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,802

- Thread starter
- Admin
- #2

- Feb 14, 2012

- 3,802

$\sqrt[3]{4(b^3+c^3)}\ge b+c$.

From $b+c>a$, it follows that $2(b+c)>a+b+c$. Thus

$\dfrac{a}{\sqrt[3]{4(b^3+c^3)}}<\dfrac{a}{b+c}<\dfrac{2a}{a+b+c}$

Therefore

$\displaystyle \sum_{\text{cyclic}}\dfrac{a}{\sqrt[3]{4(b^3+c^3)}}<\sum_{\text{cyclic}}\dfrac{2a}{a+b+c}=2$