# A Cyclic Inequality In Three Variables by Sorin Radulescu

### Solution 1

First rewrite the required inequality in a sequence of equivalent ones:

\displaystyle \begin{align} &\left[\sum_{cycl}\frac{x(y-z)^2}{xyz}\right]^3\ge 54\prod_{cycl}\frac{x(y-z)^2}{xyz}&\Leftrightarrow\\ &\left[\sum_{cycl}\frac{(y-z)^2}{yz}\right]^3\ge 54\prod_{cycl}\frac{(y-z)^2}{yz}&\Leftrightarrow\\ &\left[\sum_{cycl}\left(\frac{y}{z}+\frac{z}{y}-2\right)\right]^3\ge 54\prod_{cycl}\frac{(y-z)^2}{yz}&\Leftrightarrow\\ &\left[(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-9\right]^3\ge 54\prod_{cycl}\frac{(y-z)^2}{yz}. \end{align}

Let $s=\displaystyle \frac{1}{2}(x+y+z),$ $x=s-a,$ $y=s-b,$ $z=s-c.$ Consider a triangle with the side lengths $a,b,c$ and let $R,$ $r,$ $F$ denote its circumradius, inradius, and area, respectively. Let $r_1,$ $r_b,$ $r_c$ be its exradii. We know that

$\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{r_a}{F}+\frac{r_b}{F}+\frac{r_c}{F}=\frac{4R+r}{sr}.$

It follows that

\displaystyle\begin{align} (x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-9&=s\cdot\frac{4R+r}{sr}-9=\frac{4R-8r}{r}\\ &=4\cdot\frac{R-2r}{r}. \end{align}

\displaystyle \begin{align} &(x-y)^2+(y-z)^2(z-x)^2=(a-b)^2(b-c)^2(c-a)^2\\ &\qquad=-4r^2\left[(s^2-2R^2-10Rr+r^2)^2-4R(R-2r)^3\right]\;\text{ (Marius Dragan's)}\\ \end{align}

To continue, putting everything together we need to prove

$\displaystyle 64\frac{(R-2r)^3}{r^3}\ge\frac{54\left\{-4r^2\left[(s^2-2R^2-10Rr+r^2)^2-4R(R-2r)^3\right]\right\}}{s^2r^2}.$

\displaystyle \begin{align} &8(R-2r)^3s^2+27r(s^2-2R^2-10Rr+r^2)^2-27\cdot 4Rr(R-2r)^3\ge 0\,\Leftrightarrow\\ &4(R-2r)^3(2s^2-27Rr)+27r(s^2-2R^2-10Rr+r^2)^2\ge 0.\\ &2s^2-27Rr=2s^2-27\frac{abc}{4F}\cdot\frac{F}{s}=\frac{8s^3-27abc}{4s}=\frac{(a+b+c)^3-27abc}{4s}\ge 0, \end{align}

by the AM-GM inequality. On the other hand, $R-2r\ge 0$ is the famous Euler's inequality.

### Solution 2

As i the previous solution, we arrive at

$\displaystyle \left[\sum_{cycl}\left(\frac{y}{z}+\frac{z}{y}-2\right)\right]^3\ge 54\prod_{cycl}\frac{(y-z)^2}{yz}$

but proceed differently. WLOG, assume $x\ge y\ge z.$ Let $\displaystyle \frac{x}{y}=a\ge 1,$ $\displaystyle \frac{y}{z}=b\ge 1.$ We have $\displaystyle \frac{x}{z}=ab$ and

$\displaystyle \sum_{cycl}\left(\frac{y}{z}+\frac{z}{y}-2\right)=a+\frac{1}{a}-2+b+\frac{1}{b}-2+ab+\frac{1}{ab}-2,\\ \displaystyle \prod_{cycl}\frac{(y-z)^2}{yz}=\prod_{cycl}\frac{(y-z)^2}{z^2}=(a-1)^2(b-1)^2\left(\frac{1}{ab}-1\right)^2.$

Let $a+b=s, ab=p.$ The inequality is then rewrites as

\displaystyle\begin{align}\left(s+\frac{s}{p}+p+\frac{1}{p}-6\right)^3&\ge 54(a-1)^2(b-1)^2\left(\frac{1}{ab}-1\right)^2\\ &=54(1-s+p)^2\frac{(1-p)^2}{p^2}.\end{align}

or,

$\displaystyle \left[s(p+1)+p^2-6p+1\right]^3\ge 54(1-s+p)^2(1-p)^2.$

Define function $f:\;[2\sqrt{p},\infty)\to\mathbb{R}$ by

$f(s)= \left[s(p+1)+p^2-6p+1\right]^3- 54(1-s+p)^2(1-p)^2.$

Then

$f'(s)=3(p+1)[s(p+1)-p^2-6p+1]^2+108p(1-p)^2(1-s+p)\gt 0.$

Hence

\displaystyle\begin{align}f(s)&\ge f(2\sqrt{p})\\ &=[2\sqrt{p}(p+1)+p^2-6p+1]^3-54p(1-2\sqrt{p}+p)^2(1-p)^2\\ &=[2\sqrt{p}(p+1)-4p+p^2-2p+1]^3-54p(1-\sqrt{p})^4(1-p)^2\\ &=[2\sqrt{p}(p+1-2\sqrt{p})+(p-1)^2]^3-54p(1-\sqrt{p})^4(1-p)^2\\ &=\{(\sqrt{p}-1)^2[2\sqrt{p}+(\sqrt{p}+1)^2]\}^3-54p(1-\sqrt{p})^4(1-p)^2\\ &=(\sqrt{p}-1)^6[(p+4\sqrt{p}+1)^3-54p(1+\sqrt{p})^2]\ge 0. \end{align}

Suffice it to prove that $(p+4\sqrt{p}+1)^3-54p(1+\sqrt{p})^2\ge 0.$ Now note that $\displaystyle p+1=(\sqrt{p})^2+1\ge\frac{(\sqrt{p}+1)^2}{2},$ implying

\displaystyle \begin{align} p+4\sqrt{p}+1 &\ge\frac{(\sqrt{p}+1)^2}{2}+4\sqrt{p}\\ &=\frac{(\sqrt{p}+1)^2}{2}+2\sqrt{p}+2\sqrt{p}\\ &\ge 3\sqrt[3]{\frac{(\sqrt{p}+1)^2}{2}\cdot 2\sqrt{p}\cdot 2\sqrt{p}}\\ &=3\sqrt[3]{2p\cdot (\sqrt{p}+1)^2}. \end{align}

Thus, we see that

$\displaystyle (p+4\sqrt{p}+1)^3\ge 27\cdot 2p\cdot (\sqrt{p}+1)^2=54p(1+\sqrt{p})^2.$

### Acknowledgment

Marian Dinca has kindly communicated to me this problem by Sorin Radulescu, along with two solutions of his.