A Cyclic Inequality In Three Variables by Sorin Radulescu
Problem
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.
Cyclic inequalities in three variables
- ab + bc + ca does not exceed aa + bb + cc
- A Cyclic Inequality in Three Variables $\left(\displaystyle\frac{a^3}{b^2(5a+2b)}+\frac{b^3}{c^2(5b+2c)}+\frac{c^3}{a^2(5c+2a)}\ge\frac{3}{7}\right)$
- A Cyclic Inequality in Three Variables II $\left(\displaystyle\frac{10a^3}{3a^2+7bc}+\frac{10b^3}{3b^2+7ca}+\frac{10c^3}{3a^2+7ab}\ge a+b+c\right)$
- A Cyclic Inequality in Three Variables III $\left(\displaystyle\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge 2\right)$
- A Cyclic Inequality in Three Variables IV $\left(\displaystyle 2\sum_{cycl}(a+b)^3+5\sum_{cycl}a^3\ge 21\sum_{cycl}a^2b\right)$
- A Cyclic Inequality in Three Variables V $\left(\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{\sqrt{3(a^2+b^2+c^2)}\cdot (a+b+c)}{ab+bc+ca}\right)$
- A Cyclic Inequality in Three Variables VI $\left(\displaystyle \frac{2(a+b+c)}{abc}\ge\sum_{cycl}\left(\sqrt{\frac{a+b}{2ac}}+\sqrt{\frac{2a}{c(a+b)}}\right)\right)$
- A Cyclic Inequality in Three Variables VII $\left(\displaystyle\sum_{cycl}x\sqrt{x^2z^2+y^4}\ge\sqrt{2}\sum_{cycl}xz\sqrt{yz}\right)$
- A Cyclic Inequality in Three Variables VIII $\left(\displaystyle\sum_{cycl}(x^2+y^2)z+\sum_{cycl}\frac{xy}{(x+y)^2}\ge 27xyz\right)$
- A Cyclic Inequality in Three Variables IX $\left(\displaystyle 9\left(\sum_{cycl}\frac{x^2}{y^2}\right)^2\ge 8\left(\sum_{cycl}\frac{x}{y}\right)\left(\sum_{cycl}\frac{x^3}{y^3}-3\right)\right)$
- A Cyclic Inequality in Three Variables X $\left(\displaystyle\sum_{cycl}\frac{1}{(a+1)^3}+4\sum_{cycl}\frac{1}{(a+1)^4}\ge\frac{9}{8}\right)$
- A Cyclic Inequality in Three Variables XI $\left(\displaystyle\sum_{cycl}\frac{1}{(a^2-ab+b^2)(b^2-bc+c^2)}\le\sum_{cycl}\frac{1}{a^4}\right)$
- A Cyclic Inequality in Three Variables XII $\left(\displaystyle\left(\sum_{cycl}\frac{1}{(a^2-ab+b^2)^6}\right)^2\le 3\sum_{cycl}\left(\frac{a+b}{a^2+b^2}\right)^{24}\right)$
- A Cyclic Inequality in Three Variables XIII $\left(\displaystyle\sum_{cycl}\frac{a^2+b^2}{a+b}+11\sum_{cycl}\frac{ab}{a+b}\gt 6\sum_{cycl}\sqrt{ab}\right)$
- A Cyclic Inequality in Three Variables XIV $\left(\displaystyle\sum_{cycl}\frac{xy}{xy+y^2+zx}\le 1\right)$
- A Cyclic Inequality in Three Variables XV $\left(\displaystyle \frac{a(a^2+b^2)}{a^3+b^3}+\frac{b(b^2+c^2)}{b^3+c^3}+\frac{c(c^2+a^2)}{c^3+a^3}\leq \sqrt{\frac{a}{b}}+\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{a}}\right)$
- A Cyclic Inequality in Three Variables XVI $\left(\displaystyle \sum_{cycl}|(a+b)(1-ab)|\lt\frac{3}{2}+\sum_{cycl}a^2+\frac{1}{2}\sum_{cycl}a^4\right)$
- A Cyclic Inequality in Three Variables XVII $\left(\displaystyle \left(\sum_{cycl}\frac{x^2}{y^2}\right)^5 \ge 9\left(\sum_{cycl}\frac{x^3}{y^2z}\right)\left(\sum_{cycl}\frac{x}{\sqrt{yz}}\right)\left(\sum_{cycl}\frac{y}{z}\right)\right)$
- A Cyclic Inequality in Three Variables XVIII $\left(\displaystyle \left(\sum_{cycl}\sqrt{ab}\right)^6 \le 27\prod_{cycl}(a^2+ab+b^2)\right)$
- A Cyclic Inequality in Three Variables XIX $\left(\displaystyle \frac{x}{(y+z)^3}+\frac{y}{(z+x)^3}+\frac{z}{(x+y)^3}\ge\frac{27}{8(x+y+z)^2}\right)$
- A Cyclic Inequality in Three Variables XX $\left(\displaystyle 5\sum_{cycl}\sqrt{ab}\le\sum_{cycl}\sqrt[4]{(a+4b)(2a+3b)(3b+2a)(4a+b)}\le 5\right)$
- A Cyclic Inequality in Three Variables XXI $\left(\displaystyle \frac{abc}{7\sqrt{7}}\le\prod_{cycl}\frac{a^2-ab+b^2}{\sqrt{a^2+5ab+b^2}}\right)$
- A Cyclic Inequality in Three Variables XXII $\left(\displaystyle \sum_{cycl}\frac{a^3}{a^2+ab+b^2}\ge\frac{a+b+c}{3}\right)$
- A Cyclic Inequality in Three Variables XXIII $\left(\displaystyle 3(a^2+b^2+c^2)^2\ge 8abc(a+b+c)+\sum_{cycl}(a^2+b^2-c^2)^2\right)$
- A Cyclic Inequality in Three Variables XXIV $\left(\displaystyle \sum_{cycl} \frac{a^2b^2 (1+a^2)(1+b^2)}{(1+a)(1+b)}\geq 4(3-2\sqrt{2})abc(a+b+c)\right)$
- A Cyclic Inequality in Three Variables XXV $\left(\displaystyle \sum_{cycl} (a-\sqrt{ab}+b)^2\cdot\sum_{cycl}(a^2-ab+b^2)^2\ge 9a^2b^2c^2\right)$
- A Cyclic Inequality in Three Variables And One More $\left(\displaystyle \left(\sum_{cycl}x^{2m+2}\right)\cdot\left(\sum_{cycl}\frac{1}{(x+y)^{2m+2}}\right)\ge\frac{9}{4^{m+1}}\right)$
- Dorin Marghidanu's Cyclic Inequality in Three Variables $\left(\displaystyle\sum_{cycl}\sqrt{a^2-ab+b^2}\sqrt{b^2-bc+c^2}\ge a^2+b^2+c^2\right)$
- Dorin Marghidanu's Cyclic Inequality in Three Variables II
$\left(\displaystyle\sum_{cycl}\frac{ab}{(a+c)(b+c)} \ge\frac{3}{4}\right)$
Dorin Marghidanu's Cyclic Inequality in Three Variables III
$\left(\displaystyle \frac{a^3}{b^2+c}+\frac{b^3}{c^2+a}+\frac{c^3}{a^2+b}\gt \frac{3}{2}(abc)^{\frac{2}{3}}\right)$
- Leo Giugiuc's Cyclic Inequality in Three Variables $\left(\displaystyle a^2+b^2+c^2\ge 3\sqrt[3]{\frac{1}{4}(a-b)^2(b-c)^2(c-a)^2}+ab+bc+ca\right)$
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- A Cyclic Inequality in Three Variables with a Variable Hierarchy $\left(\displaystyle 2(x^2y + y^2z + z^2x+xyz) \ge (x+y) (y+z) (z+x)\right)$
- Dan Sitaru's Cyclic Inequality In Three Variables $\left(\displaystyle \frac{(5a+b)(5b+c)(5c+a)}{27(a+8c)(b+8a)(c+8b)}\geq \frac{8abc}{(5a+4b)(5b+4c)(5c+a)}\right)$
- Dan Sitaru's Cyclic Inequality in Three Variables V $\left(\displaystyle (x+y+z)^2\le \sum_{cycl}\sqrt{(x^2+xy+y^2)(y^2+yz+z^2)}\right)$
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- Problem 11867 from the American Mathematical Monthly $\displaystyle \left(\left(\frac{a^2}{a^2-ab+b^2}\right)^{\frac{1}{4}} + \left(\frac{b^2}{b^2-bc+c^2}\right)^{\frac{1}{4}} + \left(\frac{c^2}{c^2-ca+a^2}\right)^{\frac{1}{4}} \le 3\right)$
- An Inequality from the 1967 IMO Shortlist $\left(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{a^8+b^8+c^8}{a^3b^3c^3}\right)$
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- A Cyclic Inequality from the 6th IMO, 1964 $\left(\displaystyle \sum_{cycl}a^2(b+c-a)\le 3abc\right)$
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- Trigonometric Inequality with Integrals $\left(\small{\Omega(a,b)+\Omega(b,c)+\Omega(c,a)\le \sqrt{2}(a^2+b^2+c^2),\;\Omega(a,b)=\int_a^{2a}\int_b^{2b}|\sin (x-y)\cos (x+y)-\sin (x+y)|dxdy}\right)$
- A Cyclic Inequality in Three Variables by Uche E. Okeke $\left(\displaystyle \frac{(a+b)(b+c)(c+a)}{2}\ge abc+ \frac{(ab+bc+ca)^2}{a+b+c}\right)$
- Problem 4142 From Crux Mathematicorum $\displaystyle\left(\Bigr(1+\frac{a^2+b^2+c^2}{ab+bc+ca}\Bigr)^{\frac{(a+b+c)^2}{a^2+b^2+c^2}}\leq \Bigr(1+\frac{a}{b}\Bigr)\Bigr(1+\frac{b}{c}\Bigr)\Bigr(1+\frac{c}{a}\Bigr)\right)$
- Two-Sided Inequality by Dorin Marghidanu $\left(\displaystyle \sqrt{2}\le\sum_{cycl}\frac{b+c}{a+\sqrt{2(b^2+c^2)}}\le 2\right)$
- A Cyclic Inequality In Three Variables by Sorin Radulescu $\left(\displaystyle \left[\sum_{cycl}x(y-z)^2\right]^3\ge 54\prod_{cycl}x(y-z)^2\right)$
- Problem 1 from the 2017 Canada MO $\left(\displaystyle\left(\frac{a}{b-c}\right)^2+\left(\frac{b}{c-a}\right)^2+\left(\frac{c}{a-b}\right)^2\gt 2\right)$
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- Dan Sitaru's Cyclic Inequality in Three Variables VIII $\left(\displaystyle (xy+yz+zx)\sum_{cycl}\sqrt{x^2+xy+y^2}\le 3\sqrt{\prod_{cycl}(x^2+xy+y^2)}\right)$
- Hadamard's Determinant Inequalities and Applications I $\left((2-a-b-c+abc)^2\le (a^2+2)(b^2+2)(c^2+2)\right)$
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- A Cyclic Inequality in Three Variables XXVI $\left(\displaystyle \sum_{cycl}\frac{(x+y)^4+1}{(x+y)^6+1}\le\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right)$
- Dorin Marghidanu's Inequality with Powers and Reciprocals $\left(\displaystyle \sum_{cycl}\frac{a}{a^2bc+b^4+c^4}\le\frac{1}{abc}\right)$
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