# Dan Sitaru's Cyclic Inequality in Three Variables II

### Solution 1

We prove that $\displaystyle \sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}}=1+\frac{1}{a}-\frac{1}{a+1}.\,$ Indeed, by squaring,

\displaystyle \begin{align} &1+\frac{1}{a^2}+\frac{1}{(a+1)^2}=1+\frac{1}{a^2}+\frac{1}{(a+1)^2}+\frac{2}{a}-\frac{2}{a+1}-\frac{2}{a(a+1)}\\ &0=2\Bigr(\frac{1}{a}-\frac{1}{a+1}-\frac{1}{a(a+1)}\Bigr)\\ &0=\frac{a+1-a-1}{a(a+1)}\Leftrightarrow 0=0. \end{align}

It follows that

\displaystyle\begin{align} &\sum \sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}}=\sum \Bigr(1+\frac{1}{a}-\frac{1}{a+1}\Bigr)\\ &\qquad\qquad=\sum \Bigr(1+\frac{a+1-a}{a(a+1)}\Bigr)\\ &\qquad\qquad=3+\sum \frac{1}{a^2+a}\overbrace{\geq}^{Bergstrom} 3+\frac{9}{\sum a^2+\sum a}\\ &\qquad\qquad=3+\frac{9}{(\sum a^2)-2\sum ab+3}\\ &\qquad\qquad=3+\frac{9}{9+3-2\sum ab}=3+\frac{9}{12-2(ab+bc+ca)}. \end{align}

### Solution 2

We have

\begin{align}a^2(a+1)^2+a^2+(a+1)^2&=a^2(a+1)^2+2a(a+1)+1\\ &=(a^2+a+1)^2. \end{align}

Hence, the required inequality is equivalent to

\displaystyle \begin{align} \sum_{cycl}\frac{1}{a(a+1)}&\ge\frac{9}{12-2(ab+bc+ca)}\;\Leftrightarrow\\ &\sum_{cycl}\frac{1}{a}\ge\frac{9}{12-2(ab+bc+ca)}+\sum_{cycl}\frac{1}{a+1}. \end{align}

We'll show that $\displaystyle \frac{3}{2}\ge \frac{9}{12-2(ab+bc+ca)}\,$ which is equivalent to $3\ge ab+bc+ca\,$ and the latter is well known consequence of the constraint $a+b+c=3.$

Thus, suffice it to prove that $\displaystyle \sum_{cycl}\frac{1}{a}\ge \frac{3}{2}+\sum_{cycl}\frac{1}{a+1},\,$ which is $\displaystyle \sum_{cycl}\left(\frac{1-a}{2}\right)\left(\frac{1}{a(a+1)}+\frac{1}{a}\right)\ge 0.$

But the functions $\displaystyle \frac{1-a}{2}\,$ and $\displaystyle \frac{1}{a(a+1)}+\frac{1}{a}\,$ are both decreasing, hence, by Chebyshev's inequality,

\displaystyle \begin{align} &\sum_{cycl}\left(\frac{1-a}{2}\right)\left(\frac{1}{a(a+1)}+\frac{1}{a}\right)\\ &\qquad\qquad\ge\frac{1}{3}\left(\frac{3-a-b-c}{2}\right)\left(\frac{1}{a(a+1)}+\frac{1}{a}\right)=0. \end{align}

### Solution 3

By Minkowski's inequality,

$\displaystyle \sum_{cycl}\sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}}\ge\sqrt{9+\left(\sum_{cycl}\frac{1}{a}\right)^2+\left(\sum_{cycl}\frac{1}{a+1}\right)^2}.$

Now note that

$\displaystyle\mathbf{\circ}\qquad\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{1}{a+b+c}=\frac{9}{3}=3,\\ \displaystyle\mathbf{\circ}\qquad\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{1}{a+b+c+3}=\frac{9}{6}=\frac{3}{2}$

Thus,

$\displaystyle \sum_{cycl}\sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}}\ge\sqrt{9+9+\frac{9}{4}}=\frac{9}{2}.$

Suffice it to prove that

$\displaystyle \frac{9}{2}\ge\frac{9}{12-2(ab+bc+ca)}+3,$

or equivalently,

$\displaystyle \frac{3}{2}\ge\frac{9}{12-2(ab+bc+ca)},$

i.e.,

$36-6(ab+bc+ca)\ge 12,$

or

$ab+bc+ca\le 3,$

which is true because

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

### Solution 4

$y=x^{-2}\,$ being a convex function,

$\displaystyle \frac{1+x^{-2}}{2}\ge\left(\frac{1+x}{2}\right)^{-2},$

i.e.,

$\displaystyle 1+\frac{1}{x^2}\ge\frac{8}{(1+x)^2},$

implying

$\displaystyle 1+\frac{1}{x^2}+\frac{1}{(1+)^2}\ge\frac{9}{(1+x)^2}.$

Thus

$\displaystyle \sum_{cycl}\sqrt{1+\frac{1}{a^2}+\frac{1}{(1+a)^2}}\ge\sum_{cycl}\sqrt{\frac{9}{(a+1)^2}}=3\sum_{cycl}\frac{1}{a+1}.$

By the AM-HM inequality, $\displaystyle \sum_{cycl}\frac{1}{1+a}\ge\frac{\displaystyle 9}{\displaystyle \sum_{cycl}(a+1)}=\frac{3}{2}\,$ so that

(1)

$\displaystyle \sum_{cycl}\sqrt{1+\frac{1}{a^2}+\frac{1}{(1+a)^2}}\ge 3\sum_{cycl}\frac{1}{a+1}\ge\frac{9}{2}.$

Further, $\displaystyle \sum_{cycl}ab\le\frac{\displaystyle \left(\sum_{cycl}a\right)^2}{3}=3,\,$ such that $\displaystyle 12-2\sum_{cycl}ab\ge 6\,$ and

(2)

$\displaystyle \frac{\displaystyle 9}{\displaystyle 12-2\sum_{cycl}ab}+3\le\frac{9}{6}+3=\frac{9}{2}.$

From (1) and (2),

$\displaystyle \frac{\displaystyle 9}{\displaystyle 12-2\sum_{cycl}ab}+3\le\frac{9}{2}\le\sum_{cycl}\sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}}.$

### Solution 5

First,

$\displaystyle 1+\frac{1}{a^2}+\frac{1}{(a+1)^2} = \frac{(a^2+a+1)^2}{a^2(a+1)^2}$

so that

$\displaystyle \sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}} = \frac{a^2+a+1}{a(a+1)}=1+\frac{1}{a(a+1)}.$

Second,

$\displaystyle (a+b+c)^2=9=a^2+b^2+c^2+2(ab+bc+ca)$

so that

$\displaystyle 12-2(ab+bc+ca)=a^2+b^2+c^2+3.\,$

So, the inequality reduces to

$\displaystyle \sum_{cycl}\left(1+\frac{1}{a(a+1)}\right)\ge\frac{9}{3+a^2+b^2+c^2}+3,$

which simplifies further to

$\displaystyle \sum_{cycl}\frac{1}{a(a+1)}\ge\frac{9}{a^2+a+b^2+b+c^2+c}.$

For simplicity,let $x=a(a+1),\,$ $y=b(b+1),\,$ $z=c(c+1).\,$ The required inequality becomes

$\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}.$

Multiplying by $(x+y+z)\,$ we get

$\displaystyle 1+\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+1+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}+1\ge 9,$

or,

$\displaystyle \left(\frac{y}{x}+\frac{x}{y}\right)+\left(\frac{z}{y}+\frac{y}{z}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\ge 6$

which is true because, for $w\gt 0,\,$ $\displaystyle w+\frac{1}{w}\ge 2.$

### Solution 6

$\displaystyle \sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}} =\frac{a^2+a+1}{a(a+1)} \geq\frac{3}{a+1} ~\text{(AM-GM to the numerator)},$

implying

\displaystyle\begin{align} LHS\,&\geq 3\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\\ &\geq \frac{27}{(a+1)+(b+1)+(c+1)}\\ &= \frac{9}{2}~\text{(AM-HM)}. \end{align}

From power-mean inequality,

$\displaystyle a^2+b^2+c^2 \geq \frac{(a+b+c)^2}{3} =3,$

implying

$2(ab+bc+ca)=(a+b+c)^2-(a^2+b^2+c^2) \leq 3^2-3=6,$

and

$\displaystyle RHS\leq \frac{9}{12-6}+3=\frac{9}{2}.$

Thus,

$\displaystyle LHS\geq\frac{9}{2}\geq RHS.$

### Acknowledgment

Dan Sitaru has kindly shared a problem from his book Math Accent, with a solution of his (Solution 1) mailed on a LaTeX file, which I appreciate greatly. He also posted the problem at the CutTheKnotMath facebook page where it gathered some comments. Solution 2 is by Leo Giugiuc; Solution 3 is by Tri Nitrotoluen and independently by Subham Jaiswal. Ravi Prakash came up with a solution very close to Solution 1. Solution 4 is by Abdur Rahman; Solution5 is by Mike Lawler; Solution 6 is by Amit Itagi..