# Inequality 101 from the Cyclic Inequalities Marathon

### Solution 1

WLOG, $a\ge b\ge c,\,$ implying $\displaystyle \frac{1}{(b+1)(c+1)}\ge\frac{1}{(c+1)(a+1)}\ge\frac{1}{(b+1)(c+1)}.$ It follows by Chebyshev's inequality that

(1)

$\displaystyle LHS\ge\frac{1}{3}\left(\sum_{cycl}a^5\right)\left(\sum_{cycl}\frac{1}{(b+1)(c+1)}\right).$

By Chebyshev's inequality,

(2)

$\displaystyle \sum_{cycl}a^5\ge\frac{1}{3^4}\left(\sum_{cycl}a\right)^5=\frac{1}{81}.$

Further, by Bergstrom's inequality,

(3)

\displaystyle\begin{align} \sum_{cycl}\frac{1}{(b+1)(c+1)}&\ge\frac{\displaystyle 9}{\displaystyle \sum_{cycl}ab+2\sum_{cycl}a+3}=\frac{9}{\displaystyle \sum_{cycl}ab+5}\\ &\ge\frac{9}{\displaystyle \frac{16}{3}}=\frac{27}{16} \end{align}

because $\displaystyle \sum_{cycl}ab\le\frac{1}{3}\left(\sum_{cycl}a\right)^2.\,$ With (1)-(3),

$\displaystyle LHS\ge \frac{1}{3}\cdot\frac{1}{81}\cdot\frac{27}{16}=\frac{1}{144}.$

### Solution 2

By Hölder's inequality,

\displaystyle\begin{align}1&=(a+b+c)^5\\ &\le\left(\sum_{cycl}\frac{a^5}{(b+1)(c+1)}\right)\left(\sum_{cycl}(b+1)(c+1)\right)(1+1+1)^3. \end{align}

It follows that

$\displaystyle \left(\sum_{cycl}\frac{a^5}{(b+1)(c+1)}\right)(ab+bc+ca+5)3^3\ge 1.$

The latter implies

$\displaystyle \left(\frac{(a+b+c)^2}{3}+5\right)\left(\sum_{cycl}\frac{a^5}{(b+1)(c+1)}\right)3^3\ge 1.$

And, finally,

$\displaystyle \sum_{cycl}\frac{a^5}{(b+1)(c+1)}\ge\frac{1}{3^3}\cdot\frac{3}{16}=\frac{1}{144}.$

Equality is attained at $\displaystyle a=b=c=\frac{1}{3}.$

### Remark

The latter solution appears to equally well tackle another inequality:

Let $a,b,c\,$ be positive real numbers, subject to $a+b+c=1.\,$ Prove that

$\displaystyle \sum_{cycl}\frac{a^5}{(a+1)(b+1)}\ge\frac{1}{144}.$

### Solution 3

We use the inequality variant

(1)

$\displaystyle \left(\frac{1}{3} \left(a^p+b^p+c^p\right)\right)^{1/p}\geq \left(\frac{1}{3} \left(a^q+b^q+c^q\right)\right)^{1/q}\;,\; p \geq q$

Let $p=5, q=1$:

$\displaystyle \frac{\sqrt[5]{a^5+b^5+c^5}}{\sqrt[5]{3}}\geq \frac{1}{3} (a+b+c)$,

so

$\displaystyle a^5+b^5+c^5\geq \frac{1}{81}$

On the other hand $\displaystyle \frac{1}{27} (a+b+c+3)^3=\frac{64}{27}\geq (a+1) (b+1) (c+1)$

Expanding the lhs:

$\displaystyle lhs=\frac{a^6+a^5+b^6+b^5+c^6+c^5}{(a+1) (b+1) (c+1)}\geq \frac{27}{64}\left( \frac{1}{81}+a^6+b^6+c^6\right)$

Let $p=6, q=1$ in (1),

$\displaystyle a^6+b^6+c^6\geq \left(\frac{\sqrt[6]{3}}{3}\right)^6=\frac{1}{243}$

So

$\displaystyle lhs \geq \frac{27}{64} \left(\frac{1}{81}+\frac{1}{243}\right)=\frac{1}{144}$

### Acknowledgment

Dan Sitaru has kindly emailed me a copy of the collection RMM CYCLIC INEQUALITIES MARATHON 101-200. The problem is due to George Apostolopoulos. I copy two proofs: Solution 1 by Soumava Chakraborty, Solution 2 by Soumitra Mandal. Solution 3 is by N. N. Taleb.

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