# An Inequality and Its Modifications

### Problem

### Solution

Note that from

$\displaystyle\begin{align}\sum_{cycl}(4a-3)^2 &= 16\sum_{cycl}a^2+27-24\sum_{cycl}a\\ &=128abc-24(a+b+c) \end{align}$

it follows that $128abc-24(a+b+c)\ge 0,\,$ or,

$\displaystyle\frac{128}{24}\ge\frac{a+b+c}{abc},$

i.e.,

$\displaystyle\frac{16}{3}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}.$

Now, by the AM-GM inequality, $\displaystyle a^3+b^3+\frac{27}{64}\ge 3\sqrt[3]{\frac{27a^3b^3}{64}}=\frac{9ab}{4}.\,$ Thus,

$\displaystyle\begin{align} E&=\sum_{cycl}\frac{1}{a^3+b^3+\displaystyle\frac{27}{64}}\\ &\le\sum_{cycl}\frac{4}{9ab}\\ &\le\frac{4}{9}\cdot\frac{16}{3}\\ &=\frac{64}{27}. \end{align}$

However, $\displaystyle E\left(\frac{3}{4},\frac{3}{4},\frac{3}{4}\right)=\frac{64}{27}\,$ such that $E\,$ attains its maximum at $a=b=c=\displaystyle\frac{3}{4}.$

### Modification

Assume $a,b,c,\;$ are positive real numbers such that

$\displaystyle a^2+b^2+c^2+\frac{3}{4}=12abc.$

Prove that

$\displaystyle \frac{1}{a^3+b^3+\displaystyle\frac{1}{8}}+\frac{1}{b^3+c^3+\displaystyle\frac{1}{8}}+\frac{1}{c^3+a^3+\displaystyle\frac{1}{8}}\le 8.$

### Generalization

Assume $a,b,c,\;$ are positive real numbers and $p\,$ a positive constant such that

$\displaystyle a^2+b^2+c^2+3p^2=\frac{6}{p}abc,$

Prove that

$\displaystyle \frac{1}{a^3+b^3+p^3}+\frac{1}{b^3+c^3+p^3}+\frac{1}{c^3+a^3+p^3}\le\frac{1}{p^3}.$

Equality is only achieved for $a=b=c=p.$

### Acknowledgment

The problem above (from the Romanian Mathematical Magazine, #JP037, posed December 12, 2016) has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru, along with the solution by Kevin Soto Palacios (Peru). The problem has been proposed by Iuliana Trașcă and Neculai Stanciu (Romania).

Kunihiko Chikaya has posted (June 26, 2012) a similar problem (Modification) at the artofproblemsolving.com forum, followed by a slew of original solutions. On December 13, 2016, Kunihiko Chikaya has shared on the facebook a generalization (Generalization)

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