<s>[INST] Olympiad Trigonometric Inequality 
Let $R$ and $r$ be the circumradius and inradius of $\triangle ABC$.
  Prove that $$\frac { \cos { A }  }{ { \sin   }^{ 2 }A } +\frac { \cos { B }  }{ { \sin   }^{ 2 }B } +\frac { \cos { C }  }{ { \sin   }^{ 2 }C } \ge \frac { R }{ r }$$ 

I am not able to get a solution to this inequality. Any help would be  appreciated.
Thank you.
 [/INST] It seems the following.
Let $a$, $b$, and $c$ be the respective sides of the triangle $\triangle ABC$, $S$ be its area and $p=(a+b+c)/2$ be its semiperimeter. Then $r=S/p$ and $R=abc/4S$. Hence $R/r=abcp/4S^2$. Moreover, 
$4S^2=b^2c^2\sin^2 A=c^2a^2\sin^2 B=a^2b^2\sin^2 C$.
Hence the initial inequality is equivalent to 
$b^2c^2\cos A+ c^2a^2\cos B+ a^2b^2\cos C\ge abcp$.
But 
$\cos A=\frac{b^2+c^2-a^2}{2bc}$, 
$\cos B=\frac{c^2+a^2-b^2}{2ca}$,
$\cos C=\frac{a^2+b^2-c^2}{2ab}$.
Substituting, we obtain that the initial inequality is equivalent to 
$a^3b+b^3a+a^3c+c^3a+b^3c+c^3b\ge 2(a^2bc+ab^2c+abc^2),$
which is true by Muirhead inequality, or directly by an equivalent inequality 
$(a-b)^2(a+b)c+(b-c)^2(b+c)a+(c-a)^2 (c+a) b\ge 0$.
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