# An Inequality with Finite Sums

### Solution 1

We'll prove a more general result. If $a,x\ge 0,\;$ then, the binomial theorem,

$(a+x)^k=a^k+k\cdot a^{k-1}x+(positive\;terms)\ge a^{k-1}(a+kx),$

with equality for $k=1\;$ or, for all integer $k\ge 1\;$ when $x=0.\;$ Hence,

$\displaystyle\sum_{k=1}^{n}\frac{(a+x)^k}{a+xk}\ge \sum_{k=1}^{n}a^{k-1}=\frac{a^n-1}{a-1}.$

The equality takes place for $n=1\;$ or $x=0.\;$ The stated result is obtained with $a=2.$

### Solution 2

\displaystyle\begin{align} \sum_{k=1}^{n}\frac{(2+x)^k}{2+xk} &= \sum_{k=1}^{n}\frac{2^k(1+\displaystyle\frac{x}{2})^k}{2(1+\displaystyle\frac{x}{2}k)}\\ &\ge\sum_{k=1}^{n}2^{k-1}\cdot 1\;\;(by\;Bernoulli's\; inequality)\\ &=\frac{2^n-1}{2-1}\\ &=2^n-1. \end{align}

### Solution 3

For any $k\ge 1,\;$ the function $\displaystyle f_{k}(x)=\frac{(2+x)^k}{2+xk}\;$ is not decreasing on $[0,\infty ),\;$ so that $\displaystyle \sum_{k=1}^{n}f_{k}(x)\ge\sum_{k=1}^{n}f_{k}(0)=2^n-1.$

### Acknowledgment

Dorin Marghidanu has posted the above problem at the CutTheKnotMath facebook page, along with his solution (Solution 1). He later added Solution 2 by Kunihiko Chikaya. Solution 3 is by Leo Giugiuc. Practically the same proof has been posted by Marian Dinca.