# Nth Root of 1 plus x not greater than 1 plus x over n

## Theorem

Let $x \in \R_{>0}$ be a (strictly) positive real number.

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then:

$\sqrt [n] {1 + x} \le 1 + \dfrac x n$

## Proof

$\left({1 + y}\right)^n \ge 1 + n y$

which holds for:

$y \in \R$ where $y > -1$
$n \in \Z_{\ge 0}$

Thus it holds for $y \in \R_{> 0}$ and $n \in \Z_{> 0}$.

So:

 $\displaystyle 1 + n y$ $\le$ $\displaystyle \left({1 + y}\right)^n$ $\displaystyle \leadsto \ \$ $\displaystyle 1 + n \frac x n$ $\le$ $\displaystyle \left({1 + \frac x n}\right)^n$ substituting $y = \dfrac x n$ $\displaystyle \leadsto \ \$ $\displaystyle 1 + x$ $\le$ $\displaystyle \left({1 + \frac x n}\right)^n$ $\displaystyle \leadsto \ \$ $\displaystyle \sqrt [n] {1 + x}$ $\le$ $\displaystyle 1 + \dfrac x n$ Root is Strictly Increasing

$\blacksquare$