# A Few Variants of a Popular Inequality And a Generalization

### Solution, Problem 3

The inequality is equivalent to

$\displaystyle \sum_{cycl}\frac{n}{n+(b+c)^2}\le\frac{3n}{n+4}.$

Applying some algebra,

\displaystyle\begin{align} &\sum_{cycl}\frac{n}{n+(b+c)^2}\le\frac{3n}{n+4}&\Leftrightarrow\\ &\sum_{cycl}\frac{n+(b+c)^2-(b+c)^2}{n+(b+c)^2}\le\frac{3n}{n+4}&\Leftrightarrow\\ &\sum_{cycl}\left(1-\frac{(b+c)^2}{n+(b+c)^2}\right)\le\frac{3n}{n+4}&\Leftrightarrow\\ &\sum_{cycl}\frac{(b+c)^2}{n+(b+c)^2}\ge\frac{12}{n+4}. \end{align}

Now, with Bergstrom's inequality,

\displaystyle\begin{align}\sum_{cycl}\frac{(b+c)^2}{n+(b+c)^2}&\ge\frac{\displaystyle 4\left(\sum_{cycl} a\right)^2}{\displaystyle 3n+\sum_{cycl}(b+c)^2}\\ &=\frac{\displaystyle 4\left(\sum_{cycl}a^2+2\sum_{cycl}bc\right)}{\displaystyle n\sum_{cycl}bc+2\sum_{cycl}a^2+2\sum_{cycl}bc}\\ &=\frac{\displaystyle 4\left(\sum_{cycl}a^2+2\sum_{cycl}bc\right)}{\displaystyle 2\sum_{cycl}a^2+(n+2)\sum_{cycl}bc}. \end{align}

Thus, suffice it to prove that

$\displaystyle \frac{\displaystyle 4\left(\sum_{cycl}a^2+2\sum_{cycl}bc\right)}{\displaystyle 2\sum_{cycl}a^2+(n+2)\sum_{cycl}bc}\ge \frac{3n}{n+4}.$

The latter is equivalent to

\displaystyle \begin{align} &(n+4)\sum_{cycl}a^2+(2n+8)\sum_{cycl}bc\ge 6\sum_{cycl}a^2+(3n+6)\sum_{cycl}bc &\Leftrightarrow\\ &(n-2)\sum_{cycl}a^2\ge(n-2)\sum_{cycl}bc, \end{align}

which (after the division by $n-2)$ is well known. Equality is attained for $a=b=c=1.$

### An Application

Assume $\displaystyle \alpha+\beta+\gamma=\frac{\pi}{2}.$ Then

$\displaystyle \sum_{cycl}\frac{1}{\displaystyle 2+3\left(\tan\beta+\tan\gamma\right)^2}\le\frac{1}{2}.$

Indeed, we know that $\displaystyle \sum_{cycl}\tan\beta\tan\gamma=1.$ Then, with $x=\sqrt{3}\tan\alpha,$ $y=\sqrt{3}\tan\beta,$ $z=\sqrt{3}\tan\gamma,$ the inequality reduces to

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

### Acknowledgment

The above follows an article În Legătură Cu o Problemă Dată În Azerbaijan 2016, TST, BMO by Marin Chirciu (Romanian Mathematical Magazine, n 18, March 2017, 19-21). I am grateful to Dan Sitaru for providing me a copy of the publication.

Another variant of this problem has been discussed elsewhere.