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:
- $\paren {1 + y}^n \ge 1 + n y$
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:
\(\ds 1 + n y\) | \(\le\) | \(\ds \paren {1 + y}^n\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1 + n \frac x n\) | \(\le\) | \(\ds \paren {1 + \frac x n}^n\) | substituting $y = \dfrac x n$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1 + x\) | \(\le\) | \(\ds \paren {1 + \frac x n}^n\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sqrt [n] {1 + x}\) | \(\le\) | \(\ds 1 + \dfrac x n\) | Root is Strictly Increasing |
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $13$