# A Cyclic Inequality in Three Variables IV

### Solution 1

The required inequality is equivalent to $\displaystyle 9\sum_{cycl}a^3+6\sum_{cycl}ab^2\ge 15\sum_{cycl}a^2b.\,$ Using the AM-GM inequality,

\displaystyle\begin{align} 6a^3+6b^2a &\ge 12a^2b\\ 6b^3+6c^2b &\ge 12b^2c\\ 6c^3+6a^2c &\ge 12c^2a. \end{align}

Summing the three up gives

(A)

$\displaystyle 6\sum_{cycl}a^3+6\sum_{cycl}ab^2\ge 12\sum_{cycl}a^2b.$

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

\displaystyle\begin{align} 3(a^3+b^3+c^3) &= (a^3+a^3+b^3)+(b^3+b^3+c^3)+(c^3+c^3+a^3)\\ &\ge 3a^2b+3b^2c+3c^2a. \end{align}

Adding this to (A) proves the required inequality.

### Solution 2

The required inequality reduces to,

$\displaystyle 3\sum_{cycl}a^3+2\sum_{cycl}a^2b+2\sum_{cycl}ab^2\ge 7\sum_{cycl}a^2b.$

By the AM-GM inequality,

\displaystyle\begin{align} a^3+a^2b+ab^2 &\ge 3a^2b\\ b^3+b^2c+bc^2 &\ge 3b^2c\\ c^3+c^2a+ca^2 &\ge 3c^2a. \end{align}

Summing up we get

(1)

$\displaystyle\sum_{cycl}a^3+\sum_{cycl}a^2b+\sum_{cycl}ab^2\ge 3\sum_{cycl}a^2b.$

On the other hand,

\displaystyle\begin{align} 3(a^3+b^3+c^3) &= (a^3+a^3+b^3)+(b^3+b^3+c^3)+(c^3+c^3+a^3)\\ &\ge 3a^2b+3b^2c+3c^2a. \end{align}

So that $\displaystyle\sum_{cycl}a^3\ge\sum_{cycl}a^2b.\,$ adding this to twice (1) gives

$\displaystyle 3\sum_{cycl}a^3+2\sum_{cycl}a^2b+2\sum_{cycl}ab^2\ge 7\sum_{cycl}a^2b$

as expected.

### Solution 3

By the AM-GM inequality,

\displaystyle\begin{align} a^3+a^3+ab^2+ab^2 &\ge 4a^2b\\ b^3+b^3+bc^2+bc^2 &\ge 4b^2c\\ c^3+c^3+ca^2+ca^2 &\ge 4c^2a. \end{align}

Summing up gives $\displaystyle\sum_{cycl}a^3+\sum_{cycl}ab^2\ge 2\sum_{cycl}a^2b.\,$ Denote the left-hand side $X.\,$ Further, the required inequality reduces to $\displaystyle 9\sum_{cycl}a^3+6\sum_{cycl}ab^2\ge 15\sum_{cycl}a^2b\,$ which can be written as $\displaystyle \sum_{cycl}a^3+2X\ge 5\sum_{cycl}a^2b.\,$ It will be proved correct if $\displaystyle \sum_{cycl}a^3\ge \sum_{cycl}a^2b.\,$ But this is true due to the Rearrangement inequality.

### Acknowledgment

Dan Sitaru has kindly posted the above problem at the CutTheKnotMath facebook page, followed by three solutions. Solution 1 is by Kevin Soto Palacios (Peru); Soution 2 is by Soumava Chakraborty (India); Solution 3 is by Seyran Ibrahimov (Azerbaijan).