An Inequality and Its Modifications


An Inequality and Its Modifications


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,




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 $a=b=c=\displaystyle\frac{3}{4}.$


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

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

Prove that

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


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 $a=b=c=p.$


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 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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