An Inequality In Triangle with Sines of Half-Angles

Problem

An Inequality In Triangle with Sines of Half-Angles

Solution 1

Observe that $\displaystyle\frac{a}{b+c}=\frac{\displaystyle\sin\frac{A}{2}}{\displaystyle\cos\frac{B-C}{2}}\ge\sin\frac{A}{2}.\,$ Thus suffice it to p rove that

$\displaystyle\sum_{cycl}\sin\frac{A}{2}\le\sum_{cycl}\frac{a}{b+c}\le\sum_{cycl}\frac{2a}{(\sqrt[3]{b}+\sqrt[3]{c})(\sqrt[3]{b^2}+\sqrt[3]{c^2})}.$

To prove this, we'll verify that

$\displaystyle 2(b+c)\ge (\sqrt[3]{b}+\sqrt[3]{c})(\sqrt[3]{b^2}+\sqrt[3]{c^2}).$

The latter is equivalent to

$\displaystyle 2(b+c)\ge (b+c)+\sqrt[3]{bc}(\sqrt[3]{b}+\sqrt[3]{c})$

and, further, to $\displaystyle (b+c)\ge \sqrt[3]{bc}(\sqrt[3]{b}+\sqrt[3]{c}),\,$ i.e.

$\displaystyle (\sqrt[3]{b}+\sqrt[3]{c})(\sqrt[3]{b^2}-\sqrt[3]{bc}+\sqrt[3]{c^2})\ge \sqrt[3]{bc}(\sqrt[3]{b}+\sqrt[3]{c})$

and, finally, to

$\displaystyle (\sqrt[3]{b}+\sqrt[3]{c})(\sqrt[3]{b}-\sqrt[3]{c})^2\ge 0,$

which is obviously true.

Solution 2

Let $\sqrt[3]{b}=x\,$ and $\sqrt[3]{c}=y.\,$ Then,

$\displaystyle\begin{align} (\sqrt[3]{b}+\sqrt[3]{c})(\sqrt[3]{b^2}+\sqrt[3]{c^2})&= (x+y)(x^2+y^2)\\ &=x^3+y^3+xy(x+y)\\ &\le x^3+y^3+x^3+y^3=2(x^3+y^3)\\ &=2(b+c). \end{align}$

It follows that

$\displaystyle\begin{align} 2\sum_{cycl}\frac{a}{(\sqrt[3]{b}+\sqrt[3]{c})(\sqrt[3]{b^2}+\sqrt[3]{c^2})}&\ge 2\sum_{cycl}\frac{a}{2(b+c)}\\ &=\sum_{cycl}\frac{a}{b+c}\\ &=\sum_{cycl}\frac{2R\sin A}{2R(\sin B+\sin C)}\\ &=\sum_{cycl}\frac{\displaystyle2\sin\frac{A}{2}\cos\frac{A}{2}}{\displaystyle2\cos\frac{A}{2}\cos\frac{B-C}{2}}\\ &=\sum_{cycl}\frac{\displaystyle\sin\frac{A}{2}}{\displaystyle\cos\frac{B-C}{2}}. \end{align}$

Now, since $0\lt B,C\lt\displaystyle\frac{\pi}{2},\,$ $\displaystyle -\frac{\pi}{4}\lt\frac{B-C}{2}\lt\frac{\pi}{4},\,$ so that $\displaystyle\frac{1}{\sqrt{2}}\lt\cos\frac{B-C}{2}\le 1,\,$ and, therefore,

$\displaystyle \sum_{cycl}\frac{\displaystyle\sin\frac{A}{2}}{\displaystyle\cos\frac{B-C}{2}}\ge\sum_{cycl}\sin\frac{A}{2}$

which comletes the proof.

Acknowledgment

Kevin Soto Palacios (Peru) has kindly posted at the CutTheKnotMath facebook page his solution to a problem by Dan Sitaru. The problem was originally published in the Romanian Mathematical Magazine. Solution 2 is by Soumava Chakraborty.

 

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