# Kunihiko Chikaya's Inequality with Parameter

### Solution

We rewrite the right-hand side as $\displaystyle \sum_{cycl}\left(\sqrt{a}\cdot\sqrt{\frac{a}{pa+b}}\right).$ By Cauchy's inequality,

$\displaystyle \left[\sum_{cycl}\left(\sqrt{a}\cdot\sqrt{\frac{a}{pa+b}}\right)\right]^2\le\sum_{cycl}a\cdot\sum_{cycl}\frac{a}{ap+b}.$

Hence, suffice it to show that

\displaystyle \begin{align} &\frac{a}{ap+b}+\frac{b}{bp+c}+\frac{c}{cp+a}\le\frac{3}{p+1}&\Leftrightarrow\\ &\frac{1}{\displaystyle p+\frac{b}{a}}+\frac{1}{\displaystyle p+\frac{c}{b}}+\frac{1}{\displaystyle p+\frac{a}{c}}\le\frac{3}{p+1}. \end{align}

Define $\displaystyle x=\frac{b}{a},$ $\displaystyle y=\frac{c}{b},$ $\displaystyle z=\frac{a}{c}.$ Then $xyz=1$ and the required inequality becomes

$\displaystyle \frac{1}{p+x}+\frac{1}{p+y}+\frac{1}{p+z}\le\frac{3}{p+1}\le 1,$

which is equivalent to

$(p^2-2p)(x+y+z)+(2p-1)(xy+yz+zx)\ge 3p^2-3.$

$p^2-2p\ge 0,$ $2p-1\ge 0$ and, by the AM-GM inequality, $x+y+z\ge 3$ and $xy+yz+zx\ge 3,$ implying

\begin{align}(p^2-2p)(x+y+z)&+(2p-1)(xy+yz+zx)\\&\ge 3(p^2-2p)+3(2p-1)=3p^2-3.\end{align}

### Acknowledgment

Leo Giugiuc has kindly posted that problem of Kunihiko Chikaya at the CutTheKnotMath facebook page and later communicated to me his solution.

Copyright © 1996-2018 Alexander Bogomolny

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