# An Inequality with Cyclic Sums on Both Sides

### Solution 1

Observe that, by the AM-GM inequality,

\displaystyle \begin{align} 4\frac{a^9}{b^6c^2}+\frac{b^9}{c^6a^2}+\frac{c^9}{a^6b^2}&\ge 6\left(\frac{a^{36}}{b^{24}c^8}\cdot\frac{b^9}{c^6a^2}\cdot\frac{c^9}{a^6b^2}\right)^{\frac{1}{6}}\\&=6\left(\frac{a^{28}}{b^{17}c^5}\right)^{\frac{1}{6}},\\ \frac{a^9}{b^6c^2}+4\frac{b^9}{c^6a^2}+\frac{c^9}{a^6b^2}&\ge 6\left(\frac{b^{28}}{c^{17}a^5}\right)^{\frac{1}{6}},\\ \frac{a^9}{b^6c^2}+\frac{b^9}{c^6a^2}+4\frac{c^9}{a^6b^2}&\ge 6\left(\frac{c^{28}}{a^{17}b^5}\right)^{\frac{1}{6}}. \end{align}

Adding up and dividing by $6\,$ yields the required inequality.

### Solution 2

The required inequality is equivalent to

$\displaystyle \sum_{cycl}\frac{a^9}{b^6c^2}\ge\sqrt[6]{abc}\sum_{cycl}\sqrt{\frac{a^9}{b^6c^2}}.$

By the AM-GM inequality,

(A)

$\displaystyle \frac{a^9}{b^6c^2}+\frac{b^9}{c^6a^2}+\frac{c^9}{a^6b^2}\ge 3\sqrt[3]{abc}.$

By the Cauchy-Schwarz inequality,

(B)

$\displaystyle 3\left(\frac{a^9}{b^6c^2}+\frac{b^9}{c^6a^2}+\frac{c^9}{a^6b^2}\right)\ge \left(\sqrt{\frac{a^9}{b^6c^2}}+\sqrt{\frac{b^9}{c^6a^2}}+\sqrt{\frac{c^9}{a^6b^2}}\right)^2.$

Multiplying (A) and (B) we obtain

$\displaystyle \left(\frac{a^9}{b^6c^2}+\frac{b^9}{c^6a^2}+\frac{c^9}{a^6b^2}\right)^2\ge \sqrt[3]{abc}\left(\sqrt{\frac{a^9}{b^6c^2}}+\sqrt{\frac{b^9}{c^6a^2}}+\sqrt{\frac{c^9}{a^6b^2}}\right)^2$

which is the same as

$\displaystyle \sum_{cycl}\frac{a^9}{b^6c^2}\ge\sqrt[6]{abc}\sum_{cycl}\sqrt{\frac{a^9}{b^6c^2}}.$

### Solution 3

Let $a=x^6,\,$ $b=y^6,\,$ $c=z^6.\,$ The inequality can be written as

$\displaystyle \sum_{cycl}\frac{x^{54}}{y^{36}z^{12}}\ge\sum_{cycl}\frac{x^{28}}{y^{17}z^5}.$

This is equivalent to

$\displaystyle \sum_{cycl}\frac{x^{90}z^{24}}{y^{36}y^{36}z^{36}}\ge\sum_{cycl}\frac{x^{64}y^{19}}{z^{31}}{x^{36}y^{36}z^{36}}.$

Or,

$\displaystyle \sum_{cycl}x^{90}z^{24}\ge\sum_{cycl}x^{64}y^{19},$

### Solution 4

Rewriting

$\displaystyle LHS=\frac{a^4b^{15}+a^{15}c^4+b^4c^{15}}{a^6b^6c^6},\;RHS=\frac{a^2b^{15/2}+a^{15/2}c^2+b^2c^{15/2}}{a^{17/6}b^{17/6}c^{17/6}}.$

By the power-mean inequality,

$\displaystyle \frac{a^4b^{15}+a^{15}c^4+b^4c^{15}}{a^6b^6c^6}\ge\frac{\left(a^2b^{15/2}+a^{15/2}c^2+b^2c^{15/2}\right)^2}{3a^6b^6c^6}.$

Thus, suffice it to prove that

$\displaystyle \frac{\left(a^2b^{15/2}+a^{15/2}c^2+b^2c^{15/2}\right)^2}{3a^6b^6c^6}\ge\frac{a^2b^{15/2}+a^{15/2}c^2+b^2c^{15/2}}{a^{17/6}b^{17/6}c^{17/6}},$

or that

$\displaystyle \frac{(a^2b^{15/2}+a^{15/2}c^2+b^2c^{15/2})(a^2b^{15/2}+a^{15/2}c^2+b^2c^{15/2}-3a^{19/6}b^{19/6}c^{19/6})}{3a^6b^6c^6}\ge 0.$

The latter inequality is equivalent to

$\displaystyle a^2b^{15/2}+a^{15/2}c^2+b^2c^{15/2}\ge 3a^{19/6}b^{19/6}c^{19/6}$

which is true, since by the AM-GM inequality,

$\displaystyle \frac{a^2b^{15/2}+a^{15/2}c^2+b^2c^{15/2}}{3}\ge \sqrt[3]{a^{19/2}b^{19/2}c^{19/2}}=a^{19/6}b^{19/6}c^{19/6}.$

### Acknowledgment

Dan Sitaru has kindly posted a problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page by. Solution 1 is by Ravi Prakash (India); Solution 2 is by Kevin Soto Palacios (Peru); Solution 3 is by Amit Itagi (USA); Solution 4 is by N. N. Taleb.