<s>[INST] Prove that $1-x^n \le \dfrac{n}{n+1}\cdot \dfrac{1}{x}$ if $0Hy guys. I'm trying to prove this inequality for any $n,k \in \mathbb{N}$
$$1- x^n \le \dfrac{n}{n+1}\cdot \dfrac{1}{x^k}$$
Where $0<x<1$.
My solution attempt was to use induction over $k$
For case $k=1$, we have to prove that
$$1- x^n \le \dfrac{n}{n+1}\cdot \dfrac{1}{x}$$
I thought that maybe an induction on $ n $ would solve it now, but I wondered if there was another way less laborious. Can someone help me?
 [/INST] WTS $  x ( 1  - x^n) \leq \frac{n}{n+1} $
We will show something stronger, namely $ x (  1 - x^n ) \leq \frac{ n } { (n+1) \sqrt[n] {n+1}}$.
This follows by applying weighted AM-GM:
$$ \frac{n}{n+1} = \frac{ (nx^n) \times 1 + ( 1 - x^n) \times n } { n+1} \geq \sqrt[n+1] {nx^n ( 1-x^n)^n}.$$
So $ x ( 1 - x^n) \leq \sqrt[n]{\frac{n^{n+1} } { n(n+1)^{n+1}} }= \frac{n}{(n+1) \sqrt[n]{n+1}}.$
Equality holds when $nx^n = 1 - x^n \Rightarrow  x = \frac{ 1 } { \sqrt[n] { n+1} } $.

Note: The inequality $ x^k ( 1 - x^n) \leq \frac{n}{n+1}$ can also be done in a similar manner. Try it!
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