# An Inequality In Triangle and Without

### Solution 1

By Hölder's inequality,

\displaystyle\begin{align} \left(\sum_{3\,times}(R+r)^5\right)\left(\sum_{cycl}a^5\right)&(1+1+1)(1+1+1)(1+1+1)\\ &\ge [(Ra+ra)+(Rb+rb)+(Rc+rc)]^5. \end{align}

It follows that

(A)

$\displaystyle 3(R+r)^5\left(\sum_{cycl}a^5\right)\cdot 27\ge [(Ra+rb)+(Rb+rc)+(Rc+ra)]^5.$

On the other hand, by the AM-GM inequality,

(B)

$\displaystyle [(Ra+rb)+(Rb+rc)+(Rc+ra)]^5\left(\sum_{cycl}\frac{1}{(Ra+rb)^5}\right)\ge 729.$

because

$\displaystyle [(Ra+rb)+(Rb+rc)+(Rc+ra)]^5\ge 3^5\sqrt[3]{\left((Ra+rb)(Rb+rc)(Rc+ra)\right)^5},$

whereas

$\displaystyle \sum_{cycl}\frac{1}{(Ra+rb)^5}\ge 3\frac{1}{\sqrt[3]{\left((Ra+rb)(Rb+rc)(Rc+ra)\right)^5}},$

The product of (A) and (B) is exactly the required inequality.

### Solution 2

(1)

$\displaystyle \sum_{cycl}\frac{1}{(Ra+rb)^5}\ge \frac{(1+1+1)^6}{\displaystyle \sum_{cycl}(Ra+rb)^5}.$

Now,

\displaystyle\begin{align} \sum_{cycl}(Ra+rb)^5 &\lt (Ra+rb+Rb+rc+Rc+ra)^5\\ &=(R\sum_{cycl}a+r\sum_{cycl}a)^5=(R+r)^5\left(\sum_{cycl}a\right)^5. \end{align}

It follows that

(2)

$\displaystyle \frac{1}{\displaystyle \sum_{cycl}(Ra+rb)^5}\gt \frac{1}{\displaystyle(R+b)^5\left(\sum_{cycl}a\right)^5}.$

Combining (1) and (2),

$\displaystyle \frac{1}{\displaystyle \sum_{cycl}(Ra+rb)^5}\gt \frac{(1+1+1)^6}{\displaystyle \sum_{cycl}(Ra+rb)^5}.$

Thus, using Chebyshev's inequality,

\displaystyle\begin{align} LHS &= (R+r)^5\left(\sum_{cycl}a^5\right)\left(\sum_{cycl}\frac{1}{(Ra+rb)^5}\right)\\ &\gt\frac{\displaystyle 3^6\left(\sum_{cycl}a^5\right)}{\displaystyle\left(\sum_{cycl}a\right)^5}\ge\frac{\displaystyle 3^6\cdot\frac{1}{3^4}\left(\sum_{cycl}a\right)^5}{\displaystyle\left(\sum_{cycl}a\right)^5}\\ &=9=RHS. \end{align}

The inequality is strict.

### Acknowledgment

Kevin Soto Palacios (Peru) has kindly posted at the CutTheKnotMath facebook page his solution to a problem he credits to Dan Sitaru. Solution 2 is by Soumava Chakraborty.

As Kevin solution shows, the inequality has little to do with the triangle or its associated radii. It is true in a rectified formulation:

Let $a, b, c,s,t\,$ be positive real numbers. Prove that

$\displaystyle (s+t)^5\left(\sum_{cycl}a^5\right)\left(\sum_{cycl}\frac{1}{(sa+tb)^5}\right)\ge 9.$

I find this a most cleverly conceived occurrence of the Red Herring.