Elementary Bernoulli-Type Inequality

Question

What is an alternate way to demonstrate the inequality $$1+\frac{x}{n+1}\leq\left(1+\frac{1}{n}\right)^x$$ where $n$ is a natural number more than $0$ and $x$ is strictly between $0$ and $1$?

Seeing how Bernoulli's Inequality implies $$\left(1+\frac{1}{n}\right)^x\leq 1+\frac{x}{n}\,,$$ this is a fairly tight inequality.

I was able to prove this inequality with elementary calculus by considering the function $$\left(1+\frac{1}{n}\right)^x-\frac{x}{n+1}-1$$ on $(0,1)$ and appealing to the fairly tight inequality $$\frac{1}{n}\geq \ln\left(1+\frac{1}{n}\right)\geq\frac{1}{n+1}\,.$$ I was hoping somebody could derive this inequality more directly however.

Answer 1

$$ \begin{align} \left(1+\frac1n\right)^x &=\left(1-\frac1{n+1}\right)^{-x}\tag1\\ &\ge\left(1-\frac{x}{n+1}\right)^{-1}\tag2\\[3pt] &\ge1+\frac{x}{n+1}\tag3 \end{align} $$ Explanation:
$(1)$: reciprocal of reciprocal
$(2)$: reciprocal of the Bernoulli inequality for $0\le x\le1$
$(3)$: $1\ge1-x^2$

Answer 2

Since you know that $$ \frac{1}{n+1}\leq \ln\left(1+\frac{1}{n}\right) \tag{1} $$ you can rewrite $$ \left(1+\frac{1}{n}\right)^x = e^{x\ln\left(1+\frac{1}{n}\right)} \operatorname*{\geq}_{(1)} e^{\frac{x}{n+1}} \geq 1+\frac{x}{n+1} $$ the last inequality being the standard $1+u \leq e^u$.

Note that this implies the inequality holds for all $x\in(0,\infty)$, not only in $(0,1)$.