# A Cyclic Inequality with Powers 2 through 7

### Solution 1

By the power-mean inequality, for $n\ge m,$

$\displaystyle \left(\frac{a^n+b^n}{2}\right)^{\frac{1}{n}}\ge\left(\frac{a^m+b^m}{2}\right)^{\frac{1}{m}}$

which can be rewritten as

$\displaystyle \frac{a^n+b^n}{a^m+b^m}\ge \left(\frac{a^m+b^m}{2}\right)^{\frac{n}{m}-1},$

implying, by the AM-GM inequality,

$\displaystyle \frac{a^n+b^n}{a^m+b^m} \ge \left(a^m+b^m\right)^{\frac{n}{m}-1}\ge (ab)^{\frac{n-m}{2}}.$

It follows that

\displaystyle\begin{align} \frac{(a^7+b^7)^3}{(a^4+b^4)(a^5+b^5)(a^6+b^6)}&=\frac{a^7+b^7}{a^4+b^4}\cdot\frac{a^7+b^7}{a^5+b^5}\cdot\frac{a^7+b^7}{a^6+b^6}\\ &\ge (ab)^{\frac{7-4}{2}} (ab)^{\frac{7-5}{2}} (ab)^{\frac{7-6}{2}}=(ab)^3. \end{align}

Again, with the AM-GM inequality,

\displaystyle\begin{align}\sum_{cycl}\frac{(a^7+b^7)^3}{(a^4+b^4)(a^5+b^5)(a^6+b^6)}&\ge\sum_{cycl}(ab)^3\\ &\ge 3a^2b^2c^2. \end{align}

### Solution 2

For $k<7$,

\displaystyle \begin{align} \frac{a^7+b^7}{a^k+b^k}&\geq 2^{\left(1-7/k\right)}(a^k+b^k)^{\left(7/k-1\right)}~\text{(Power-mean/numerator)}\\ &\geq 2^{\left(1-7/k\right)}2^{\left(7/k-1\right)}(ab)^{\left(\frac{7-k}{2}\right)}~\text{(AM-GM)}.\\ &=(ab)^{\left(\frac{7-k}{2}\right)} \end{align}

Thus,

\displaystyle \begin{align} \sum_{cyc} \frac{(a^7+b^7)^3}{(a^4+b^4)(a^5+b^5)(a^6+b^6)}&\geq \sum_{cyc} (ab)^{\left(\frac{3}{2}+1+\frac{1}{2}\right)}\\ &=\sum_{cyc} (ab)^3\geq 3a^2b^2c^2~\text{(AM-GM)}. \end{align}

### Acknowledgment

Dan Sitaru has kindly posted at the CutTheKnotMath facebook page the above problem of his with a solution (Solution 1) by Lazaros Reggina Zachariades. The problem was previously published in the Romanian Mathematical Magazine. Solution 2 is by Amit Itagi.