# Prove $x>nδ_n$, $\forall n∈\mathbb{N}+$ for $δ_n := \sqrt[n]{x}−1$

I' trying to prove that for $x>1$, $x>nδ_n$, $\forall n∈\mathbb{N}+$ where $δ_n := \sqrt[n]{x}−1$ and $\delta_n>0$. I tried doing this by induction but got stuck on the last part. $$nδ_n=n(\sqrt[n]{x}−1)$$

1. Base case: $n=1$: $$x>nδ_n\implies x>x-1$$ Therefore, it is true for $n=1$
2. Assume true for $n=k$: $$x>k(\sqrt[k]{x}−1)$$
3. Examine case $n=k+1$: $$x>(k+1)(\sqrt[k+1]{x}−1)$$

This is where I get stuck. I don't see how you can implement the induction hypothesis from part 2. Is Proof by Induction the right method to prove this? If so, any help on continuing the proof would be appreciated.

We have $x=(1+\delta_n)^n$.Now apply the Bernoulli inequality
We'll prove that $f(x)\geq0$ for $x\geq0$, where $f(x)=\left(1+\frac{x}{n}\right)^n-x$.
$f'(x)=\left(1+\frac{x}{n}\right)^{n-1}-1\geq0$.
Thus, $f(x)\geq f(0)=0$ and we are done!
• How did you get this $f(x)$? – aL_eX Nov 13 '16 at 17:34
• $f(x)\geq0\Leftrightarrow 1+\frac{x}{n}\geq\sqrt[n]x\Leftrightarrow x\geq n\left(\sqrt[n]x-1\right)$ – Michael Rozenberg Nov 13 '16 at 17:37