# Dorin Marghidanu's Example for Radon's Inequality

### Solution 1

\displaystyle\begin{align} \frac{a^{m+n+1}}{(b^{\frac{m}{m+n}}c^{\frac{n}{m+n}})^{m+n}}&+\frac{b^{m+n+1}}{(c^{\frac{m}{m+n}}a^{\frac{n}{m+n}})^{m+n}}+\frac{c^{m+n+1}}{(a^{\frac{m}{m+n}}b^{\frac{n}{m+n}})^{m+n}}\\ &\ge \frac{(a+b+c)^{m+n+1}}{(b^{\frac{m}{m+n}}c^{\frac{n}{m+n}}+c^{\frac{m}{m+n}}a^{\frac{n}{m+n}}+a^{\frac{m}{m+n}}b^{\frac{n}{m+n}})^{m+n}}. \end{align}

By the Rearrangement Inequality,

\displaystyle\begin{align}b^{\frac{m}{m+n}}c^{\frac{n}{m+n}}+c^{\frac{m}{m+n}}a^{\frac{n}{m+n}}+a^{\frac{m}{m+n}}b^{\frac{n}{m+n}}&\le c^{\frac{m}{m+n}}c^{\frac{n}{m+n}}+a^{\frac{m}{m+n}}a^{\frac{n}{m+n}}+b^{\frac{m}{m+n}}b^{\frac{n}{m+n}}\\ &=a+b+c. \end{align}

It follows that

\displaystyle\begin{align}\frac{a^{m+n+1}}{(b^{\frac{m}{m+n}}c^{\frac{n}{m+n}})^{m+n}}&+\frac{b^{m+n+1}}{(c^{\frac{m}{m+n}}a^{\frac{n}{m+n}})^{m+n}}+\frac{c^{m+n+1}}{(a^{\frac{m}{m+n}}b^{\frac{n}{m+n}})^{m+n}}\\&\ge\frac{(a+b+c)^{m+n+1}}{(a+b+c)^{m+n}}\\ &=a+b+c. \end{align}

A combination of the two inequalities yields the required one.

### Solution 2

By the AM-GM inequality, say, $\displaystyle b^mc^n\le \left(\frac{1}{m+n}(mb+nc)\right)^{m+n}.\;$ Thus,

\displaystyle\begin{align} &\frac{a^{m+n+1}}{b^mc^n}+\frac{b^{m+n+1}}{c^ma^n}+\frac{c^{m+n+1}}{a^mb^n}\ge\\ &\;\;\;\frac{a^{m+n+1}}{\left(\frac{1}{m+n}(mb+nc)\right)^{m+n}}+\frac{b^{m+n+1}}{\left(\frac{1}{m+n}(mc+na)\right)^{m+n}}+\frac{c^{m+n+1}}{\left(\frac{1}{m+n}(ma+nb)\right)^{m+n}}. \end{align}

To this we apply Radon's inequality:

\displaystyle\begin{align} \frac{a^{m+n+1}}{\left(\frac{1}{m+n}(mb+nc)\right)^{m+n}}&+\frac{b^{m+n+1}}{\left(\frac{1}{m+n}(mc+na)\right)^{m+n}}+\frac{c^{m+n+1}}{\left(\frac{1}{m+n}(ma+nb)\right)^{m+n}}\\&\ge\frac{(a+b+c)^{m+n+1}}{\left(\displaystyle\frac{(mb+nc)+(mc+na)+(ma+nb}{m+n}\right)^{m+n}}\\ &=\frac{(a+b+c)^{m+n+1}}{(a+b+c)^{m+n}}\\ &=a+b+c. \end{align}

### Solution 3

By the weighted means inequality (which is the xtension of the regular AG-AM inequality),

\displaystyle\begin{align} \frac{a^{m+n+1}}{b^mc^n}+mb+nc &\ge (m+n+1)\cdot\sqrt[m+n+1]{\frac{a^{m+n+1}}{b^mc^n}b^mc^n}\\ &=(m+n+1)a. \end{align}

Hence,

(1)

$\displaystyle\frac{a^{m+n+1}}{b^mc^n}+mb+nc \ge (m+n+1)a.$

Similarly,

(2)

$\displaystyle\frac{b^{m+n+1}}{c^ma^n}+mc+na \ge (m+n+1)b$

and

(3)

$\displaystyle\frac{c^{m+n+1}}{a^mb^n}+ma+nb \ge (m+n+1)c.$

Adding up (1), (2), and (3) yields

$\displaystyle \sum_{cycl}\frac{a^{m+n+1}}{b^mc^n}+\sum_{cycl}(mb+nc) \ge (m+n+1)\sum_{cycl}a,$

i.e.,

$\displaystyle \sum_{cycl}\frac{a^{m+n+1}}{b^mc^n}+(m+n)\sum_{cycl}a \ge (m+n+1)\sum_{cycl}a,$

or,

$\displaystyle \sum_{cycl}\frac{a^{m+n+1}}{b^mc^n}\ge \sum_{cycl}a,$

which is the required inequality.

Equality in (1) is achieved iff

$\displaystyle\frac{a^{m+n+1}}(b^mc^n}=b=c,$

i.e., only if $\displaystyle\frac{a^{m+n+1}}{b^{m+n}}=b,\;$ implying $a=b\;$ and, therefore, $a=b=c.$

### Acknowledgment

Dorin Marghidanu has kindly posted the elegant problem above at the CutTheKnotMath facebook page. Solutions 2 and 3 are by Dorin Marghidanu.