# A Cyclic Inequality in Three Variables II

### Solution 1

Since the inequality is homogeneous, we may assume WLOG, $a^2+b^2+c^2=1.$ First using the GM-QM inequality and then Chebyshev's inequality twice, we get

\displaystyle\begin{align} \sum_{cycl}\frac{10a^3}{3a^2+7bc} &\ge \sum_{cycl}\left(\frac{10a^3}{3a^2+\displaystyle 7\left(\frac{b^2+c^2}{2}\right)}\right)\\ &=20\sum_{cycl}\frac{a^3}{6a^2+7b^2+7c^2}\\ &=20\sum_{cycl}\left(\frac{a^3}{7-a^2}\right)\\ &\ge\frac{20}{9}\left(\sum_{cycl}a\right)\left(\sum_{cycl}a^2\right)\left(\sum_{cycl}\frac{1}{7-a^2}\right)\\ &\ge\frac{20}{9}\left(\sum_{cycl}a\right)\left(\frac{9}{\displaystyle\sum_{cycl}(7-a^2)}\right)\\ &=\sum_{cycl}a, \end{align}

as desired.

### Solution 2

Since the inequality is homogeneous, we may assume WLOG, $abc=1.\,$ The inequality is then rewritten as $\displaystyle_{cycl}f(a)\ge 0,\,$ where

$\displaystyle f(x)\frac{10x^4}{3x^3+7}-x=\frac{7x(x^3-1)}{3x^3+7}.$

The function is convex, so that $f(x)\ge g(x)=\displaystyle\frac{21}{10}(x-1),\,$ which is its tangent at $x=1.\,$ Thus

$\displaystyle\sum_{cycl}f(a)\ge\sum_{cycl}g(a)=\frac{63}{10}-\frac{63}{10}=0,$

because $x+y+z\ge 3\sqrt[3]{xyz}=3.\,$ This is in fact the required inequality.

### Solution 3

By Hölder's inequality,

$\displaystyle (a+b+c)^3 \le \sum_{cycl}\frac{a^3}{3a^2+7bc}\cdot\sum_{cycl}(3a^2+7bc)\cdot\sum_{cycl}1.$

Thus, suffice it to prove that $\displaystyle\sum_{cycl}(3a^2+7bc)\le\frac{10(a+b+c)^2}{3},\,$ or, in other words, that $a^2+b^2+c^2\ge a+b+c.\;$ But then

\displaystyle\begin{align} \sum_{cycl}\frac{10a^3}{3a^2+7bc} &\ge \frac{10(a+b+c)^3}{\displaystyle 3\sum_{cycl}(3a^2+7bc)}\\ &\ge\frac{10(a+b+c)^3}{10(a+b+c)^2}\\ &=a+b+c. \end{align}

### Solution 4

We have a series of equivalent inequalities:

\displaystyle\begin{align} &\frac{10a^3}{3a^2+7bc}+\frac{10b^3}{3b^2+7ca}+\frac{10c^3}{3a^2+7ab}\ge a+b+c,\\ &\sum_{cycl}\left(\frac{10a^3}{3a^2+7bc}-a\right)\ge 0,\\ &\frac{7}{2}\sum_{cycl}\frac{a(a+b)(a-c)+a(a-b)(a+c)}{3a^2+7bc}\ge 0,\\ &\frac{7}{2}\sum_{cycl}\left[\frac{a(a-b)(a+c)}{3a^2+7bc}+\frac{a(a+b)(a-c)}{3a^2+7bc}\right]\ge 0,\\ &\frac{7}{2}\sum_{cycl}\left[\frac{a(a-b)(a+c)}{3a^2+7bc}+\frac{b(b+c)(b-a)}{3b^2+7ca}\right]\ge 0,\\ &\frac{7}{2}\sum_{cycl}(a-b)\left[\frac{a(a+c)}{3a^2+7bc}-\frac{b(b+c)}{3b^2+7ca}\right]\ge 0,\\ &\frac{7}{2}\sum_{cycl}(a-b)\left[\frac{7c(a^3-b^3)+3abc(b-a)+7c^2(a^2-b^2)}{(3a^2+7bc)(3b^2+7ca)}\right]\ge 0,\\ &\frac{7}{2}\sum_{cycl}(a-b)^2\left[\frac{7c(a^2+b^2)+7c^2(a+b)+4abc}{(3a^2+7bc)(3b^2+7ca)}\right]\ge 0.\\ \end{align}

The latter is obviously true and, so, the rest are also true.

### Acknowledgment

The problem above (from the Romanian Mathematical Magazine) has been kindly communicated to me by Dan Sitaru, along with four solutions. Solution 1 is by Anas Adlany (Morroco); Solution 2 is by Imad Zak (Lebanon); Solution 3 is by Kevin Soto Palacios (Peru); Solution 3 is by Diego Alvariz (India); Solution 4 is by Soumitra Moukherjee (India).