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For all triangle prove that $$\sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}\geq\frac{3}{2}\sqrt[9]{\frac{2r}{R}}$$

Here $R$ it's the radius of a circumcircle , $r$ is the radius of an incircle,

$\alpha$, $\beta$ and $\gamma$ are measured-angles of the triangle.

I tried Holder and more, but without success.

Replacing $9$ at $10$ gives a wrong inequality already.

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    I think you may already have tried using $4\sin\frac{A}2\sin\frac{B}2\sin\frac{C}2 = \frac{r}{R}$ to convert it fully to trigonometry? – 2017-01-20
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    @Sawarnik After using this statement we still need to prove the starting inequality. I don't see how it helps. – 2017-01-20
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    is this $$\sqrt[9]{..}$$ or $$\sqrt[3]{..}$$? – 2017-01-20
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    @Dr. Sonnhard Graubner it's $\sqrt[9]{...}$ – 2017-01-20
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    Assuming that the $r/R$ formula presented by @Sawarnik is true, then the relationship holds replacing $\frac32$ with $1.62008$. And replacing 9 by 10 does not give a wrong inequality; you still can have $\frac32$ replaced by anything up to $1.59875$. I suspect that they used $\frac32$ and a ninth root because that is somehow easier to prove. **oops** I did that mistakenly ignoring the $2$ in $2r/R$. With that $2$ present, the inequality seems to be tight. – 2017-01-20
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    @Mark Fischler After the Ravi's substitution try $y=z=1$ and you'll get a wrong inequality. – 2017-01-20

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