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

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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

From Bernoulli's Inequality:

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

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

$\blacksquare$


Sources