# A Cyclic Inequality in Three Variables XVIII

### Solution 1

First note that

1. $a^2+ab+b^2\ge\displaystyle \frac{3}{4}(a+b)^2\ge 0\,\Longleftrightarrow\,(a-b)^2\ge 0,$
2. $a,b,c\gt 0\,\Longrightarrow\,\displaystyle \frac{9}{8}\prod_{cycl}(a+b)\ge\left(\sum_{cycl}a\right)\left(\sum_{cycl}ab\right),$
3. $a,b,c\in\mathbb{R}\,\Longrightarrow\,(a+b+c)^2\ge 3(ab+bc+ca).$

It follows that

\displaystyle \begin{align} \prod_{cycl}(a^2+ab+b^2) &\ge 27\left(\frac{3}{4}\right)^3\prod_{cycl}(a+b)^2\\ &\ge 9(a+b+c)^2(ab+bc+ca)^2. \end{align}

However,

$9(a+b+c)^2(ab+bc+ca)^2\ge 27(ab+bc+ca)^3.$

Thus, suffice it to prove that

$27(ab+bc+ca)^3\ge (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^6.$

But, by the Cauchy-Schwarz inequality,

\displaystyle \begin{align} 27(ab+bc+ca)^3 &\ge 27\left(\frac{(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2}{3}\right)^3\\ &\ge (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^6. \end{align}

### Solution 2

We know that $a^2+ab+b^2\ge\displaystyle \frac{3}{4}(a+b)^2+\frac{(a-b)^2}{4}\ge\frac{3}{4}(a+b)^2.\,$ Similarly, $\displaystyle b^2+bc+c^2\ge\frac{3}{4}(b+c)^2\,$ and $\displaystyle c^2+ca+a^2\ge\frac{3}{4}(c+a)^2,\,$ implying

\displaystyle\begin{align} 27\prod_{cycl}(a^2+ab+b^2)&\ge 27\cdot\left(\frac{3}{4}\right)^3\prod_{cycl}(a+b)^2\\ &\ge 27\cdot\left(\frac{3}{4}\right)^3\frac{64}{81}\left(\sum_{cycl}a\right)^2\left(\sum_{cycl}ab\right)^2\\ &\ge 27\left(\prod_{cycl}ab\right)^3\\ &\ge \left(\sum_{cycl}\sqrt{ab}\right)^6, \end{align}

where we used the following inequalities:

1. $\displaystyle 9\prod_{cycl}(a+b)\ge 8\left(\sum_{cycl}a\right)\left(\sum_{cycl}ab\right),$
2. $(a+b+c)^2\ge 3(ab+bc+ca),$
3. $\displaystyle \frac{ab+bc+ca}{3}\ge\left(\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{3}\right)^2.$

### Solution 3

Lemma 1

For $x,y,z\ge 0,$

(1)

$\displaystyle \left(\sum_{cycl}xy\right)^3\leq \prod(x^2+xy+y^2)$

Indeed, from Hölder's inequality:

\displaystyle \begin{align} \prod_{cycl}(x^2+xy+y^2)&=(xy+y^2+x^2)(y^2+yz+z^2)(x^2+z^2+xz)\\ &\geq (\sqrt[3]{xy\cdot y^2\cdot x^2}+\sqrt[3]{y^2\cdot yz\cdot z^2}+\sqrt[3]{x^2\cdot z^2\cdot xz})^3\\ &=(xy+yz+zx)^3. \end{align}

Lemma 2

For $x,y\ge 0,$

(2)

$\displaystyle \left(\sum_{cycl}xy\right)^3\leq 27 \prod_{cycl}(x^2-xy+y^2).$

Indeed,

$(x-y)^2\geq 0\Rightarrow 2(x-y)^2\geq 0\Rightarrow 2x^2-4xy+2y^2\geq 0$

and, finally,

$3x^2-3xy+3y^2\geq x^2+xy+y^2.$

or,

$x^2-xy+y^2\geq \frac{1}{3}(x^2+xy+y^2)\,$

Similarly,

$y^2-yz+z^2\geq \frac{1}{3}(y^2+yz+z^2)\,$ and
$z^2-zx+x^2\geq \frac{1}{3}(z^2+zx+x^2).$

By multiplying the last three relationships,

(3)

$\displaystyle \prod_{cycl}(x^2+xy+y^2)\leq 27 \prod_{cycl}(x^2-xy+y^2).$

Multiplying that by (1) and (2),

\displaystyle\begin{align} \left(\sum_{cycl}xy\right)^6&\leq 27\prod_{cycl}(x^2+xy+y^2)\cdot \prod_{cycl}(x^2-xy+y^2)\\ &=27\prod_{cycl}\left((x^2+y^2)^2-x^2y^2\right)\\ &=27\prod_{cycl}(x^4+y^4+x^2y^2). \end{align}

Thus,

(4)

$\displaystyle \left(\sum_{cycl}xy\right)^6=27\prod_{cycl}(x^4+y^4+x^2y^2).$

Setting now in (4) $x=\sqrt{a};\, y=\sqrt{b};\, z=\sqrt{c},\,$ we get

$\displaystyle \left(\sum_{cycl}\sqrt{ab}\right)^6 \le 27\prod_{cycl}(a^2+ab+b^2).$

The equality holds if $a=b=c$.

### Solution 4

Case 1: Exactly one of $a,b,c=0$

With $a=0,\,$ the given inequality reduces to $27b^2c^2(b^2+bc+c^2)\ge b^3c^3,\,$ which is obvious. Cases $b=0\,$ and $c=0\,$ are handled similarly.

Case 2: At least two of $a,b,c=0$

The inequality reduces to $0\ge 0.$

Case 3: $a,b,c \gt 0$

Using Wu's inequality, $RHS\ge 27(ab+bc+ca)^3.\,$ Thus, suffice it to prove that

\begin{align} &27(ab+bc+ca)^3\ge (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^6,\,\Longleftrightarrow\\ &3(ab+bc+ca)\ge (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2,\,\Longleftrightarrow\\ &3\sum_{cycl}x^2\ge\left(\sum_{cycl}x)\right)^2, \end{align}

where $x=\sqrt{ab},\,$ $y=\sqrt{bc},\,$ $z=\sqrt{ca}.\,$ The latter is true by Hölder's inequality.

### Solution 5

We have, by the Power-Mean Inequality, $\displaystyle \left(\frac{1}{3}(\sum_{cycl}(ab)^{\frac{1}{2}}\right)^2\le\frac{1}{3}\sum_{cycl}ab.\,$ Hence

$\displaystyle LHS=\left(\sum_{cycl}(ab)^{\frac{1}{2}}\right)^6\le 3^3\left(\sum_{cycl}ab\right)^3$

and we need to prove that $\displaystyle 3^3\left(\sum_{cycl}ab\right)^3\le RHS,\,$ or

$\displaystyle \left(\sum_{cycl}a^4b^2+\sum_{cycl}a^2b^4+\sum_{cycl}a^4bc\right)-\left(\sum_{cycl}ab^2c^3+\sum_{cycl}a^2b^2c+\sum_{cycl}a^2b^2c^2\right)\ge 0,$

which is true by rearrangement.

### Acknowledgment

This problem has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru and then again by Kevin Soto Palacios, along with his solution (Solution 1). Solution 2 is by Diego Alvariz; Solution 3 is by Dan Sitaru; Solution 4 is by Soumava Chakraborty; Solution 5 is by Nassim Nicholas Taleb. The problem is by Dan Sitaru and has been previously published at the Romanian Mathematical Magazine.