# Refinement on Dan Sitaru's Cyclic Inequality In Three Variables

### Preliminaries

An earlier page dealt with a problem by Dan Sitaru:

While solving that problem, N. N. Taleb has observed (How the problem came about below) the existence of an upper bound on the right-hand side of the inequality and suggested a refinement that is the subject of the present page.

An early attempt of solving the new problem (Solution 1) relied on the graphics produced by wolframalpha. Leo Giugiuc devised Solution 2, Amit Itagi Solution 3.

### How the problem came about

First of all, we simplify the problem by replacing the variables: $a=x,\,$ $b=2y,\,$ $c=3z,\,$ which reduces the problem to

Prove that if $a,b,c\gt 0;\,abc(a+b+c)=1\,$ then:

$\displaystyle \frac{(a^2b^2+1)(b^2c^2+1)(c^2a^2+1)}{a^2b^2c^2}\geq \frac{64}{(a+b+c)^2}.$

We start with the constraint by applying the AM-GM inequality: $1\ge (abc)(a+b+c)\ge abc\cdot 3\sqrt[3]{abc}=3(abc)^{\frac{4}{3}}\,$ so that $\displaystyle abc\le\frac{1}{3^{3/4}},\,$ implying a bound for the RHS of the inequality,

$\displaystyle \frac{1}{(a+b+c)^2}=(abc)^2\le\left(\frac{1}{3^{3/4}}\right)^2=\frac{1}{3^{3/2}}.$

Thus, the above problem.

### Solution 1

We start with Amit Itagi's approach for solving the original problem (and copied from Solution 3 below). Let,

$\displaystyle x=\sqrt{\frac{ab}{c}},~2y=\sqrt{\frac{bc}{a}},~3z=\sqrt{\frac{ca}{b}}.$

Thus, the inequality and the constraint, respectively, become

$\displaystyle (a^2+1)(b^2+1)(c^2+1)\geq\frac{64abc}{3\sqrt{3}},~ab+bc+ca=1.$

Define $p=\sqrt[3]{abc}.\,$ From the constraint, $1\ge 3\sqrt[3]{a^2b^2c^2}=3\sqrt[3]{p^2},\,$ implying $\displaystyle p\in\left[0,\frac{1}{3\sqrt{3}}\right].\,$ Now, for the left-hand side,

\displaystyle\begin{align}(a^2+1)(b^2+1)(c^2+1)&=1+\sum_{cycl}a^2+\sum_{cycl}a^2b^2+a^2b^2c^2\\ &\ge 1+3\sqrt[3]{a^2b^2c^2}+3\sqrt[3]{a^4b^4c^4}+a^2b^2c^2\\ &=1+3p^2+3p^4+p^6. \end{align}

We, therefore, define the function

$\displaystyle f(p)=1+3p^2+3p^4+p^6-\frac{64}{3\sqrt{3}}p^3.$

The graph below affirms the inequality $f(p)\ge 0,\,$ for $\displaystyle p\in\left[0,\frac{1}{3\sqrt{3}}\right]\,$ if we notice that $\displaystyle f\left(\frac{1}{\sqrt{3}}\right)=0:$

### Solution 2

The smartest way is the following. Denote $\displaystyle 6x^2yz=\frac{a}{3},\,$ $\displaystyle 12xy^2z=\frac{b}{3},\,$ $\displaystyle 18xyz^2=\frac{c}{3}.\,$ Then $a+b+c=3\,$ and the required inequality becomes

$(ab+3c)(bc+3a)(ca+3b)\ge 64(abc)^{\frac{3}{2}}.$

Let's remark that, since $abc\le 1,\,$ $(abc)^{\alpha}\ge(abc)^{\beta},\,$ for $\alpha\le\beta.\,$ By the AM-GM inequality,

$ab+3c= ab+c+c+c\ge 4(abc^3)^{\frac{1}{4}}.$

Similarly we obtain two additional inequalities, with the product of the three

$(ab+3c)(bc+3a)(ca+3b)\ge 4^3(abc)^{\frac{5}{4}}\ge 64(abc)^{\frac{3}{2}},$

because $\displaystyle \frac{5}{4}\lt\frac{3}{2}.$

### Solution 3

Using the constraint, the inequality can be written as

$\displaystyle (4x^2y^2+1)(36y^2z^2+1)(9x^2z^2+1)\geq \frac{64(6xyz)^2}{3\sqrt{3}}.$

Let,

$\displaystyle x=\sqrt{\frac{ab}{c}},~2y=\sqrt{\frac{bc}{a}},~3z=\sqrt{\frac{ca}{b}}.$

Thus, the inequality and the constraint, respectively, become

$\displaystyle (a^2+1)(b^2+1)(c^2+1)\geq\frac{64abc}{3\sqrt{3}},~ab+bc+ca=1.$

Let $p=\sqrt{3}(abc)^{1/3}$. AM-GM gives

$\displaystyle 1=ab+bc+ca\geq3(abc)^{2/3}=p^2~\text{or}~1\geq p.$

The inequality can be simplified to

$\displaystyle 1+(a^2+b^2+c^2)+(a^2b^2+b^2c^2+c^2a^2)+(abc)^2-\frac{64abc}{3\sqrt{3}}\geq 0.$

\displaystyle \begin{align} LHS&\geq 1+3(abc)^{2/3}+3(abc)^{4/3}+(abc)^2-\frac{64abc}{3\sqrt{3}}~\text{(AM-GM)} \\ &= 1+p^2+\frac{p^4}{3}+\frac{p^6}{27}-\frac{64p^3}{27}\\ &=\frac{(p-1)(p-3)(p^4+4p^3+22p^2+12p+9)}{27}\geq 0, \end{align}

because $(p-1)(p-3)\ge 0\,$ due to the constraint $0\le p\le 1.$