# A Cyclic Inequality in Three Variables by Uche E. Okeke

### Solution 1

By the AM-GM inequality, applied three times,

$\displaystyle abc\le\frac{1}{8}(a+b)(b+c)(c+a).$

Hence, suffice it to show that

$\displaystyle \frac{(ab+bc+ca)^2}{a+b+c}\le\frac{3}{8}\frac{(a+b)(b+c)(c+a)}{2}.$

Using in sequence $(a+b+c)\ge 3(ab+bc+ca)$ and

$8(a+b+c)(ab+bc+ca)\le 9(a+b)(b+c)(c+a),$

we obtain

\displaystyle \begin{align} \frac{(ab+bc+ca)^2}{a+b+c}&=\frac{(a+b+c)(ab+bc+ca)^2}{(a+b+c)^2}\\ &\le \frac{(a+b+c)(ab+bc+ca)}{3}\\ &\le\frac{3}{8}(a+b)(b+c)(c+a). \end{align}

### Solution 2

Let $x=a+b,$ $y=b+c,$ $z=c+a.$ Then $x+y\gt z,$ $y+z\gt x,$ $z+x\gt y$ so that $x,y,z$ are form a triangle. For the semiperimeter $s,$ the circumradius $R$ and the inradius $r$ we have

$\displaystyle s=\sum_{cycl}a,\,a=s-y,\,b=s-z,\,c=s-x.$

Thus, the given inequaliy rewrites as

(1)

$\displaystyle \frac{xyz}{2}\ge\prod_{cycl}(s-x)+\frac{\displaystyle \left[\sum_{cycl}(s-x)(s-y)\right]^2}{s}.$

Now,

\displaystyle \begin{align} \sum_{cycl}(s-x)(s-y)&=\sum_{cycl}(s^2-s(2s-z)+yz=\sum_{cycl}(-s^2+sz+yz)\\ &=-3s^2+s\cdot (2s)+4Rr+r^2=4Rr+r^2. \end{align}

I.e.,

(2)

$\displaystyle\sum_{cycl}(s-x)(s-y)=4Rr+r^2.$

From (1),(2), it follows that the given inequality is equivalent to

\displaystyle\begin{align}&\frac{xyz}{2}\ge\frac{r^2s^2}{s}+\frac{r^2(4R+r)^2}{s}&\;\Leftrightarrow\\ &s(4Rrs)\ge 2r^2s^2+2r^2(4R+r)^2&\;\Leftrightarrow\\ &(2R-r)s^2\ge r(4R+r)^2.\end{align}

Using Gerretsen's inequality $s^2\ge 16Rr-5r^2,$ we get

\displaystyle \begin{align} &\frac{xyz}{2}\ge(2R-r)(16Rr-5r^2)\ge r(4R+r)^2&\;\Leftrightarrow\\ &(R-2r)(8R-r)\ge 0, \end{align}

which true, due to Euler's inequality $R\gt 2r.$

### Solution 3

Let $p=a+b+c,\,$ $q=ab+bc+ca,$ $r=abc.$ Then the required inequality becomes $\displaystyle pq-\frac{2q^2}{p}\ge 3r,$ whhich is equivalent to $(a^2+b^2+c^2)(ab+bc+ca)\ge 3abc(a+b+c),$ which is true because $a^2+b^2+c^2\ge ab+bc+ca,$ and $(ab+bc+ca)^2\ge 3abc(a+b+c).$

### Solution 4

Using Bergstrom's inequality,

\displaystyle \begin{align} \frac{1}{2}\prod_{cycl}(a+b)-abc &=\frac{1}{2}\sum_{cycl}c^2(a+b)\\ &=\frac{1}{2}\sum_{cycl}\frac{c^2(a+b)^2}{a+b}\ge\frac{(ab+bc+ca)^2}{a+b+c}. \end{align}

### Solution 5

The inequality is equivalent to

$\displaystyle \left(\sum_{cycl}a^2b+\sum_{cycl}ab^2\right)(\left(\sum_{cycl}a\right)\ge 2\left(\sum_{cycl}ab\right)^2,$

which reduces to

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

The latter follows by the AM-GM inequality from

\begin{align} a^3b+a^3b+a^3c+a^3c+b^3c+c^3b&\ge 6a^2bc\\ b^3c+b^3c+b^3a+b^3a+c^3a+a^3c&\ge 6b^2ca\\ c^3a+c^3a+c^3b+c^3b+a^3b+b^3a&\ge 6c^2ab. \end{align}

### Acknowledgment

The problem has been kindly communicated to me by Uche E. Okeke on October 30, 2017, along with several solutions. Solution 1 is by Hung Viet Nguyen; Solution 2 is by Samouva Chakraborty; Solution 3 is by Gheorghe Duca; Solution 4 is by Nevena Sybeva; Solution 5 is by Earl C.

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