# A Cyclic Inequality from Math Phenomenon

### Solution 1

With the AM-HM inequality,

$\displaystyle \frac{2}{\displaystyle \frac{1}{a}+\frac{1}{b+c}}\le \frac{a+b+c}{2},$

i.e., $\displaystyle\frac{a+b+c}{a(b+c)}\ge \frac{4}{a+b+c}.$ Similarly, $\displaystyle\frac{a+b+c}{b(c+a)}\ge \frac{4}{a+b+c}$ and $\displaystyle\frac{a+b+c}{c(a+b)}\ge \frac{4}{a+b+c}.$ Adding up all three yields the required inequality.

### Solution 2

Note that

$[a+(b+c)]^2\ge 4a(b+c)$
$[b+(c+a)]^2\ge 4b(c+a)$
$[c+(a+b)]^2\ge 4c(a+b)$

Using Bergstrom's inequality,

\displaystyle\begin{align}\sum_{cycl}\frac{1}{a(b+c)}&\ge\frac{(1+1+1)^2}{\displaystyle \sum_{cycl}a(b+c)}\\ &\ge\frac{9\cdot 4}{3[a+b+c]^2}=\frac{12}{(a+b+c)^2}. \end{align}

### Solution 3

The inequality being homogenoeus, WLOG, let $a+b+c=1$. Thus, the inequality can be written as

$\displaystyle \sum_{cyc}\frac{1}{a(1-a)}=\sum_{cyc}\frac{1}{a}+\sum_{cyc}\frac{1}{1-a}\geq 12.$

Using the concavity of $f(x)=1/x$ and Jensen's inequality (or effectively AM-HM),

\displaystyle \begin{align} &\frac{1}{6}\left(\sum_{cyc}\frac{1}{a}+\sum_{cyc}\frac{1}{1-a}\right)\geq\frac{6}{\sum_{cyc} a+\sum_{cyc}(1-a)} \\ \Rightarrow &\sum_{cyc}\frac{1}{a}+\sum_{cyc}\frac{1}{1-a}\geq\frac{36}{3}=12. \end{align}

### Solution 4

WLOG, $a+b+c=1.$ Let us show the stronger inequality

$\displaystyle 1/(a(1-a))+1/(b(1-b))+1/(c(1-c))\geq 27/2$

We want to find the minimum of the sum using method of Lagrange multipliers. We have restriction $a+b+c=1.$ If $a\rightarrow 0$ or $a\rightarrow 1$, clearly the sum tends to infinity.

Let $\displaystyle f(x)=1/(x(1-x))$. The condition of Lagrange method implies $f'(a)=f'(b)=f'(c)=\lambda$.

Note that

$\displaystyle f''(x)=\frac{-6x^2+6x-2}{x^3(x-1)^3}\geq 0$

on $(0,1)$, hence $f'$ is increasing and therefore injective on $(0,1)$.

Thus $\displaystyle a=b=c=\frac{1}{3}$ provides the inequality minimum.

### Solution 5

Using Bergstrom's inequality,

\displaystyle\begin{align}\sum_{cycl}\frac{1}{a(b+c)}&\ge\frac{(1+1+1)^2}{\displaystyle \sum_{cycl}a(b+c)}\\ &\ge\frac{9}{2(ab+bc+ca)}=\frac{27}{2(a+b+c)^2}, \end{align}

since, as well known, $3(ab+bc+ca)\le(a+b+c)^2.$

### Acknowledgment

The problem (with a slight modification) and Solution 1 are from Dan Sitaru's book Math Phenomenon, problem G.4. Solution 2 is by Mohamed Lamine; Solution 3 is by Amit Itagi; Solution 4 is by Bogdan Lataianu. Nevena Sybeva and Leo Giugiuc noticed that Solution 2 could have potentially led to a stronger inequality. Their remark is included as Solution 5.