# Dorin Marghidanu's Inequality in Integer Variables

### Solution

With Bernoulli's inequality, $(1+x)^a\lt 1+ax,\,$ true for $x\gt -1\,$ and $a\in (0,1],\,$ we have (since $\displaystyle \frac{1}{m},\frac{1}{n}\lt 1)$

(1)

$\displaystyle (1+n)^{\frac{1}{m}}\lt 1+\frac{n}{m}$

and

(2)

$\displaystyle (1+m)^{\frac{1}{n}}\lt 1+\frac{m}{n}.$

Then (1)$\Rightarrow\displaystyle \frac{1}{\sqrt[m]{1+n}}\gt\frac{m}{m+n}\,$ and (2)$\Rightarrow\displaystyle \frac{1}{\sqrt[n]{1+m}}\gt\frac{n}{m+n}\,$ such that

(3)

$\displaystyle \frac{m}{\sqrt[m]{1+n}}\gt\frac{m^2}{m+n}$

and

(4)

$\displaystyle \frac{n}{\sqrt[n]{1+m}}\gt\frac{n^2}{m+n}.$

Adding up (3) and (4) and using Bergstrom's inequality,

$\displaystyle \frac{m}{\sqrt[m]{1+n}}+\frac{n}{\sqrt[n]{1+m}}\gt\frac{m^2+n^2}{m+n}\ge\frac{(m+n)^2}{2(m+n)}=\frac{m+n}{2}.$

### Generalization

If $r_1,r_2,\ldots,r_n\in\mathbb{N}_{\ge 2},\,$ then

$\displaystyle \sum_{k=1}^n\frac{r_k}{\sqrt[r_k]{1+r_{k+1}}}\gt\frac{1}{2}\sum_{k=1}^nr_k,\, (r_{n+1}=r_1.)$

Indeed, as before, $\displaystyle \frac{r_k}{\sqrt[r_k]{1+r_{k+1}}}\gt\frac{r_k^2}{r_k+r_{k+1}}\,$ and applying Bergstrom's inequality, we have

$\displaystyle \sum_{k=1}^n\frac{r_k}{\sqrt[r_k]{1+r_{k+1}}}\gt\sum_{k=1}^n\frac{r_k^2}{r_k+r_{k+1}}\ge\frac{\displaystyle \left(\sum_{k=1}^nr_k\right)^2}{\displaystyle 2\sum_{k=1}^nr_k}=\frac{1}{2}\sum_{k=1}^nr_k,$

### Acknowledgment

The problem has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu, along with his solution and a generalization.