# A Cyclic Inequality in Three Variables

### Solution 1

Consider $\displaystyle f(x)=\frac{1}{x(5+2x)}+\frac{9}{49}\ln x-\frac{1}{7}.$

$\displaystyle f'(x)=\frac{(x-1)(36x^2+216x+245)}{49x^2(2x+5)^2}.$

$f'(x)\lt 0,\;$ for $x\lt 1,\;$ and $f'(x)\gt 0,\;$ for $x\gt 1.\;$ Since $f(1)=0,\;$ $f(x)\ge 0,\;$ for $x\gt 0.$

Now, let $\displaystyle x=\frac{b}{a},\;$ $\displaystyle y=\frac{c}{b},\;$ $\displaystyle z=\frac{a}{c}.\;$ $xyz=1.\;$ We have to show that $\displaystyle\sum_{cycl}g(x)\ge\displaystyle\frac{3}{7},\;$ where $g(x)=\displaystyle\frac{1}{x(5+2x)}.$

\displaystyle\begin{align} \sum_{cycl}g(x) &= \sum_{cycl}\left(f(x)-\frac{9}{49}\ln x+\frac{1}{7}\right)\\ &= \sum_{cycl}\left(f(x)\right)-\frac{9}{49}\ln xyz+\frac{3}{7}\\ &= \sum_{cycl}\left(f(x)\right)+\frac{3}{7}\\ &\ge 0+\frac{3}{7}. \end{align}

Equality is attained for $x=y=z=1,\;$ i.e., $a=b=c.$

### Solution 2

By the Cauchy-Schwarz inequality,

$\displaystyle\sum_{cycl}\frac{a^3}{b^2(5a+2b)}\sum_{cycl}\left[ab^2(5a+2b)\right]\ge (a^2+b^2+c^2)^2.$

Thus, suffice it to show that

$\displaystyle\sum_{cycl}\frac{a^3}{b^2(5a+2b)}\ge \frac{(a^2+b^2+c^2)^2}{\displaystyle\sum_{cycl}\left[ab^2(5a+2b)\right]}\ge \frac{3}{7}.$

This is equivalent to

$\displaystyle 7(a^2+b^2+c^2)^2\ge 3\left(5\sum_{cycl}(a^2b^2+2ab^3+2bc^3+2ca^3)\right),$

which can be written as

$\displaystyle 6\sum_{cycl}a^4+\sum_{cycl}a^4+14\sum_{cycl}a^2b^2\ge 15\sum_{cycl}(a^2b^2+6ab^3+6bc^3+6ca^3).$

With the AM-GM inequality, we see that

$a^4+a^4+a^4+c^4\ge 4a^3c,\\ b^4+b^4+b^4+a^4\ge 4b^3a,\\ c^4+c^4+c^4+b^4\ge 4c^3b.$

And summing up these (times $6)\;$ and $a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2,\;$ which we know is true, we obtain

$a^4+b^4+c^4\ge ab^3+bc^3+ca^3.$

### Solution 3

By Radon's Inequality, then the Cauchy-Schwarz inequality, and later, the AM-GM inequality,

\displaystyle\begin{align} \sum_{cycl}\frac{a^3}{b^2(5a+2b)}&\ge\sum_{cycl}\frac{(a+b+c)^3}{\left(\displaystyle\sum_{cycl}\sqrt{b}\cdot\sqrt{5ab+2b^2}\right)^2}\\ &\ge \frac{(a+b+c)^3}{ \left(\sqrt{(a+b+c)\displaystyle\sum_{cycl}(2b^2+5ab)}\right)^2 }\\ &\ge \frac{(a+b+c)^3}{\left(\sqrt{(a+b+c)[2(a+b+c)^2+ab+bc+ca]}\right)^2}\\ &\ge \frac{(a+b+c)^3}{\left(\sqrt{(a+b+c)[2(a+b+c)^2+\frac{1}{3}(a+b+c)^2]}\right)^2}\\ &=\frac{3}{7}. \end{align}

### Solution 4

Define $f:\;(0,\infty)\rightarrow\mathbb{R},\;$ by $f(x)=\displaystyle\frac{x^3}{2+5x}.\;$ Then

\displaystyle\begin{align} f'(x) &= \frac{3x^2(2+5x)-x^3\cdot 5}{(2+5x)^2}\\ &=\frac{6x^2+10x^3}{(2+5x)^2}\\ &\gt 0. \end{align}

Thus the function is strictly increasing. So, for any positive $x_1,x_2,x_3,\;$

$\displaystyle f\left(\sqrt[3]{x_1x_2x_3}\right)\le f\left(\displaystyle\frac{x_1+x_2+x_3}{3}\right).$

Further,

$\displaystyle f''(x) = \frac{50x^3+10x^2+4x}{(2+5x)^4}\gt 0,$

making the function convex. By Jensen's inequality then,

$\displaystyle f\left(\sqrt[3]{x_1x_2x_3}\right)\le f\left(\displaystyle\frac{x_1+x_2+x_3}{3}\right)\le \frac{1}{3}[f(x_1)+f(x_2)+f(x_3)].$

Set $\displaystyle x_1=\frac{a}{b},\;$ $\displaystyle x_2=\frac{b}{c},\;$ $\displaystyle x_3=\frac{c}{a}\;$ to obtain

$\displaystyle f\left(\sqrt[3]{\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}}\right)\le \frac{1}{3}\left[f\left(\frac{a}{b}\right)+f\left(\frac{b}{c}\right)+f\left(\frac{c}{a}\right)\right].$

Explicitly,

\displaystyle\begin{align} \frac{1}{7}=f(1)&\le \frac{1}{3}\sum_{cycl}\frac{\displaystyle \left(\frac{a}{b}\right)^3}{\displaystyle 2+\frac{5a}{b}}\\ &=\frac{1}{3}\sum_{cycl}\frac{a^3}{2b^3+5ab^2}. \end{align}

Hence, the required inequality.

### Solution 5

By the Cauchy-Schwarz inequality,

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

It remains to prove that

$\displaystyle 7(a^2+b^2+c^2)^2\ge 15(a^2b^2+b^2c^2+c^2a^2)+6(ab^3+bc^3+ca^3).$

This is equivalent to

$\displaystyle 7(a^4+b^4+c^4)\ge (a^2b^2+b^2c^2+c^2a^2)+6(ab^3+bc^3+ca^3).$

Since, $a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2,\;$ suffice it to show that

$\displaystyle a^4+b^4+c^4\ge ab^3+bc^3+ca^3.$

But this follows from summing up the inequalities below

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

The proof is complete.

### Acknowledgment

The problem above has been posted on the CutTheKnotMath facebook page by Dan Sitaru. The problem came from his book Math Storm. Solution 1 is by Imad Zak (Lebanon); Solution 2 is by Kevin Soto Palacios (Peru); Solution 3 is by Diego Alvariz (India); Solution 4 is by Dan Sitaru (Romania); Solution 5 by Hung Nguyen Viet (Vietnam). The illustraion is by Gary Davis.