# Dorin Marghidanu's Two-Sided Inequality

### Solution 1

From the problem's constraints, $a,b,c\in (0,1),$ implying $1-a,1-b,1-c\gt 0.$ We consider the expression

\begin{align}E(a,b,c) &=(1-a)(1-b)(1-c)\\ &=\sqrt{(1-a)(1-b)}\cdot\sqrt{(1-b)(1-c)}\cdot\sqrt{(1-c)(1-a)} \end{align}

and using the inequalities $\displaystyle \frac{2xy}{x+y}\le\sqrt{xy}\le\frac{x+y}{2},\,x,y\gt 0$ between the harmonic, geometric and the arithmetic means for each of the three roots, we get

$\displaystyle \prod_{cycl}\frac{2(1-a)(1-b)}{(1-a)+(1-b)}\le E(a,b,c)\le\prod_{cycl}\frac{(1-a)+(1-b)}{2}$

which is equivalent to

$\displaystyle \prod_{cycl}\frac{2[1-(a+b)+ab]}{1+c}\le E(a,b,c)\le\prod_{cycl}\frac{1+a}{2}$

or, equivalently,

$\displaystyle 8\cdot\prod_{cycl}\frac{c+ab}{1+c}\le E(a,b,c)\le\frac{1}{8}\prod_{cycl}(1+a)$

which resolves to the required inequality.

Equality happens for $1-a=1-b=1-c,$ i.e., $a=b=c=\displaystyle \frac{1}{3}.$

### Solution 2

$\displaystyle 8\cdot\prod_{cycl}(a+bc)=8\cdot\prod_{cycl}(a^2+ab+bc+ca).$

Now, by the AM-GM inequality,

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

implying $1-c^2\ge 2\sqrt{(a+bc)(b+ca)}$ and, similarly, $1-a^2\ge 2\sqrt{(c+ab)(b+ca)}$ and $1-b^2\ge 2\sqrt{(a+bc)(c+ab)}.$

Thus, $\displaystyle\prod_{cycl}(1-a^2)\ge 8\prod_{cycl}(a+bc).$ Further,

\displaystyle \begin{align} \frac{1+a}{2}&=\frac{(a+b)+(a+c)}{2}\ge\sqrt{(a+b)(a+c)}\,\Rightarrow\\ &1+a\ge 2\sqrt{(1-c)(1-b)}, \end{align}

and, similarly, $1+b\ge 2\sqrt{(1-c)(1-a)}$ and $1+c\ge 2\sqrt{(1-b)(1-a)}.$ It follows that

$\displaystyle \prod_{cycl}(1+a)^2\ge 8\prod_{cycl}(1-a^2)$

and, hence,

$\displaystyle \prod_{cycl}(1+a)^2\ge 8\prod_{cycl}(1-a)^2\ge 64\prod_{cycl}(a+bc),$

as required.

### Solution 3

$a+bc=1-b-c+bc=(1-b)(1-c),$ and, similarly, $b+ca=(1-c)(1-a)$ and $c+ab=(1-a)(1-b).$ It follows that

$64(a+bc)(b+ca)(c+ab)=64(1-a)^2(1-b)^2(1-c)^2$

and, also,

\begin{align} &64(1-a)^2(1-b)^2(1-c)^2 \le 8(1-a^2)(1-b^2)(1-c^2)\\ &\qquad\qquad\qquad =8(1-a)(1-b)(1-c)(1+a)(1+b)(1+c)&\Leftrightarrow\\ &8(1-a)(1-b)(1-c)\le (1+a)(1+b)(1+c)&\Leftrightarrow\\ &8(b+c)(c+a)(a+b)\le (2a+b+c)(2b+c+a)(2c+a+b)&\Leftrightarrow\\ &8xyz\le(x+y)(y+z)(z+x), \end{align}

where $x=a+b,$ $y=b+c,$ $z=c+a.$ With the AM-GM inequality, this is true $\displaystyle \prod_{cycl}(x+y)\ge\prod_{cycl}[2\sqrt{xy}]=8xyz.$

On the other hand,

$8(1-a)(1-b)(1-c)\le(1+a)(1+b)(1+c)$

is equivalent to

$8(1-a^2)(1-b^2)(1-c^2)\le(1+a)^2(1+b)^2(1+c)^2.$

### Solution 4

For the sake of reference, let $p=a+b+c,\,$ so that $p=1.\,$ Let $r=\sqrt[3]{abc}$ and $q=ab+bc+ca:$

$\displaystyle q=ab+bc+ca\le a^2+b^2+c^2\le \frac{1}{3}(a+b+c)^2,$

so that

(1)

$\displaystyle q\le\frac{1}{3}.$

Now note that

(2)

$a+bc=a(a+b+c)+bc=(a+b)(a+c).$

According to Schur's inequality, $2p^3+9r-7pq\ge 0,$ which in our case reads as

(3)

$9r-7q+2\ge 0.$

Now,

$\displaystyle \prod_{cycl}(1+a)^2-8\prod_{cycl}(1-a^2)=\prod_{cycl}(a+1)\cdot(9r-7q+2)\ge 0,$

according to (3). Equality is attained for $\displaystyle a=b=c=\frac{1}{3}.$

On the other hand, according to (2),

\displaystyle\begin{align} 8\prod_{cycl}(1-a^2)-64\prod_{cycl}(a+bc)&=8\prod_{cycl}-64\prod_{cycl}(a+b)^2\\ &=8\prod_{cycl}(1-a^2)-64\prod_{cycl}(1-a)^2\\ &=8\prod_{cycl}(1-a)\cdot (9r-7q+2)\ge 0 \end{align}

again due to (3). Equality is attained for $\displaystyle a=b=c=\frac{1}{3}.$

### Solution 5

Expanding each expression in the system of inequalities and using $a+b+c=1,$

\displaystyle \begin{align}P &= (a+bc)(b+ca)(c+ab) = (ab+bc+ca-abc)^2\\ &= [(a+b+c)(ab+bc+ca)-abc]^2 = [(a+b)(b+c)(c+a)]^2. \end{align}

Similarly,

\displaystyle \begin{align} Q &= (1-a^2)(1-b^2)(1-c^2)=(ab+bc+ca+1)^2 - (abc+1)^2\\ &= (ab+bc+ca - abc)\cdot (ab+bc+ca+abc+2)\\ &= [(a+b+c)(ab+bc+ca)-abc](1+a+b+ab+bc+ca+abc)\\ &= (a+b)(b+c)(c+a)(1+a)(1+b)(1+c), \end{align}

and if $R = (1+a)^2(1+b)^2(1+c)^2,$ then the given inequality can be rewritten as $64P \le 8Q \le R$ which is equivalent to

$\displaystyle 64\prod_{cycl}(a+b)^2 \le 8\prod_{cycl}(a+b)\cdot\prod_{cycl}(1+a) \le \prod_{cycl}(1+a)^2,$

where $1+a = (a+b+c) + a = (c+a) + (a+b),$ etc. We are to prove that

\displaystyle \begin{align}64\prod_{cycl}(a+b)^2 &\le 8\prod_{cycl}(a+b)\cdot\prod_{cycl}(a+b + b+c)\\ &\le\prod_{cycl}(a+b + b+c)^2, \end{align}

which are all true, because $(x+y)(y+z)(z+x) = 8xyz$ for positive reals $x, y, z,$ as Marian Dinca has already shown.

Note: I should have noticed the shortcut

\begin{align}1-a^2 &= (a+b+c)^2-a^2=(b+c)(2a+n+c)\\ &=(b+c)[(c+a)+(a+b)]. \end{align}

### Acknowledgment

Dorin Marghidanu has kindly posted the above problem at the CutTheKnotMath facebook page, along with his (Solution 1) and several additional solutions. Solution 2 is by Diego Alvariz; Solution 3 is by Marian Dinca; Solution 4 is by Imad Zak; Solution 5 is by Kunihiko Chikaya.