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.

  • $\begingroup$ I think we should change r so that all things become trigonometric $\endgroup$ – Archis Welankar Dec 20 '15 at 11:52

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

$b^2c^2\cos A+ c^2a^2\cos B+ a^2b^2\cos C\ge abcp$.


$\cos A=\frac{b^2+c^2-a^2}{2bc}$, $\cos B=\frac{c^2+a^2-b^2}{2ca}$, $\cos C=\frac{a^2+b^2-c^2}{2ab}$.

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

$(a-b)^2(a+b)c+(b-c)^2(b+c)a+(c-a)^2 (c+a) b\ge 0$.

  • 1
    $\begingroup$ Beautiful and elegant solution :) $\endgroup$ – Swapnil Das Dec 20 '15 at 13:21

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