triangle inequality

[ghenghea]

Prove that in any acute triangle : $\dfrac{\left( \sin{A}+\sin{B}+\sin{C} \right)^2}{ \sin{A} \cdot \sin{B} \cdot \sin{C} } \ge \dfrac{\left( \cos{\dfrac{A}{2}} +\cos{\dfrac{B}{2}} +\cos{\dfrac{C}{2}} \right) ^2}{ \cos{\dfrac{A}{2}} \cdot \cos{\dfrac{B}{2}} \cdot \cos{\dfrac{C}{2} }} $

[Sergic Primazon]

ghenghea в сообщении #1159983 писал(а):

Prove that in any acute triangle : $\dfrac{\left( \sin{A}+\sin{B}+\sin{C} \right)^2}{ \sin{A} \cdot \sin{B} \cdot \sin{C} } \ge \dfrac{\left( \cos{\dfrac{A}{2}} +\cos{\dfrac{B}{2}} +\cos{\dfrac{C}{2}} \right) ^2}{ \cos{\dfrac{A}{2}} \cdot \cos{\dfrac{B}{2}} \cdot \cos{\dfrac{C}{2} }} $

$$\Leftrightarrow \left (\cos \left( \frac{\alpha}{2} \right)+\cos \left( \frac{\beta}{2} \right)+\cos \left( \frac{\gamma}{2} \right) \right )^2\le \frac{p^2}{2Rr}$$

$$\left (\cos \left( \frac{\alpha}{2} \right)+\cos \left( \frac{\beta}{2} \right)+\cos \left( \frac{\gamma}{2} \right) \right )^2\le \left ( \frac{3 \sqrt3}{2}\right)^2 \le \frac{p^2}{2Rr} , ( 2p^2 \ge 27Rr )$$

So, it is true for all triangles)