# Dorin Marghidanu's Inequality in Many Variables Plus Two More

### Solution 1

The function, $f:\,(0,\infty)\to\mathbb{R},$ defined by $\displaystyle f(x)=\ln\left(p+\frac{r}{x}\right)$ is convex on $(0,\infty),$ for $\displaystyle f''(x)=\frac{r(2px+r)}{x^2(px+r)^2}\gt 0.$

$\displaystyle f\left(\frac{a_1+a_2+\ldots+a_n}{n}\right)\le\frac{f(a_1)+f(a_2)+\ldots+f(a_n)}{n}$

so that

\displaystyle \begin{align} \ln\left(p+\frac{r}{A_n[a_1,a_2,\ldots,a_n]}\right) &\le A_n\left[\ln\left(p+\frac{r}{a_1}\right),\ldots,\ln\left(p+\frac{r}{a_n}\right)\right]\Leftrightarrow\\ \ln\left(p+\frac{r}{A_n[a_1,a_2,\ldots,a_n]}\right) &\le \ln G_n\left[p+\frac{r}{a_1},p+\frac{r}{a_2},\ldots,p+\frac{r}{a_n}\right]\Leftrightarrow\\ p+\frac{r}{A_n[a_1,a_2,\ldots,a_n]} &\le G_n\left[p+\frac{r}{a_1},p+\frac{r}{a_2},\ldots,p+\frac{r}{a_n}\right]. \end{align}

Equality holds for $a_1=a_2=\ldots=a_n.$

### Solution 2

By Hölder's inequality,

$\displaystyle G_n\left[p+\frac{r}{a_1},p+\frac{r}{a_2},\ldots,p+\frac{r}{a_n}\right]\ge p+\frac{r}{G_n[a_1,a_2,\ldots,a_n]}.$

By the AM-GM inequality,

$\displaystyle p+\frac{r}{G_n[a_1,a_2,\ldots,a_n]}\ge p+\frac{r}{A_n[a_1,a_2,\ldots,a_n]}.$

### Acknowledgment

This problem has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu with a solution of his (Solution 1). Solution 2 is by Leo Giugiuc.