# A Cyclic Inequality in Triangle for Integer Powers

### Solution

\displaystyle \begin{align} \sum_{cycl}\frac{a^{n+1}}{b+c-a}-\sum_{cycl}a^n &=\sum_{cycl}\left(\frac{a^{n+1}}{b+c-a}-a^n)\right)\\ &=\sum_{cycl}\frac{a^{n+1}-a^nb-a^nc+a^{n+1}}{b+c-a}\\ &=\sum_{cycl}\frac{a^n(a-b)+a^n(a-c)}{b+c-a}\\ &=\sum_{cycl}\frac{a^n(a-b)}{b+c-a}+\sum_{cycl}{a^n(a-c)}{b+c-a}\\ &=\sum_{cycl}\frac{a^n(a-b)}{b+c-a}+\sum_{cycl}{b^n(b-a)}{c+a-b}\\ &=\small{\sum_{cycl}(a-b)\left(\frac{a^n(c+a-b)-b^n(b+c-a)}{(b+c-a)(c+a-b)}\right)}\\ &=\small{\sum_{cycl}(a-b)\left(\frac{a^nc+a^{n+1}-a^nb-b^{n+1}-b^nc+ab^n)}{(b+c-a)(c+a-b)}\right)}\\ &=\small{\sum_{cycl}(a-b)\left(\frac{a(a^n+b^n)-b(a^n+b^n)+c(a^n+b^n)}{(b+c-a)(c+a-b)}\right)}\\ &=\small{\sum_{cycl}(a-b)\left(\frac{(a-b)(a^n+b^n)+c(a^n-b^n)}{(b+c-a)(c+a-b)}\right)}. \end{align}

We now consider two cases:

$\mathbf{n\gt 0}$

\displaystyle \begin{align} &\sum_{cycl}(a-b)\left(\frac{(a-b)(a^n+b^n)+c(a^n-b^n)}{(b+c-a)(c+a-b)}\right)\\ &\qquad\qquad=\sum_{cycl}(a-b)^2\left(\frac{\displaystyle (a^n+b^n)+c\sum_{k=0}^{n-1}a^{n-1-k}b^k}{\displaystyle (b+c-a)(c+a-b)}\right)\ge 0. \end{align}

$\mathbf{n=0}$

\displaystyle \begin{align} &\sum_{cycl}(a-b)\left(\frac{(a-b)(a^n+b^n)+c(a^n-b^n)}{(b+c-a)(c+a-b)}\right)\\ &\qquad\qquad=\sum_{cycl}(a-b)\left(\frac{(a-b)\cdot 2}{(b+c-a)(c+a-b)}\right)\\ &\qquad\qquad=\sum_{cycl}(a-b)^2\left(\frac{2}{(b+c-a)(c+a-b)}\right)\ge 0. \end{align}

### Acknowledgment

The problem (from the Romanian Mathematical Magazine) has been kindly posted by Dan Sitaru at the CutTheKnotMath facebook page, Dan later emailed me his solution in a LaTeX file.