Bernoulli's inequality for rational exponents

Question

We've proved that $$(1+x)^n \geq 1+nx \quad \forall n \in \mathbb{N} \land x \geq -1$$ with induction, and next excercises are to prove $1$ and $2$: $$(1+x)^p \leq 1+px \quad \forall p \in \mathbb{Q}\cap[0,1] \land x \geq -1 \tag{1}$$ $$(1+x)^p \geq 1+px \quad \forall p \in \mathbb{Q}\cap[1,\infty) \land x \geq -1 \tag{2}$$ I was told that $(1)$ will be useful in proving $(2)$, so it's suggested to prove the $(1)$ first.

My work:
For the first one, i was able to prove that $$(1+x)^{1/n} \leq 1+\frac{x}{n} \quad \forall n \in \mathbb{N}_{+} \land x \geq -1$$ Like this: $$(1+x)^{1/n} \overset{?}{\leq} 1+\frac{x}{n}$$ $$(1+x)\overset{?}{\leq} \left(1+\frac{x}{n}\right)^n$$ $$ 1+nx \leq (1+x)^n$$ Which is true, but I could not go further.

But I was able to prove $(2)$ from $(1)$: for $q \in \mathbb{Q}\cap (0,1]$, we have that $$(1+x)^q \leq 1+qx$$ Letting $pq=1$: $$1+x \leq \left( 1+\frac{x}{p}\right)^p$$ $$1+px \leq (1+x)^p$$ Could you give me a hint to the first one?

Answer 1

For your inequality $(1)$, here is a way: let $p = \dfrac{m}n \in (0, 1)$ where $m, n \in \mathbb N$. So $m < n$, and we may write $(1)$ as $$\sqrt[n]{\color{red}{1\cdot1\cdot1\cdots1}\cdot\color{blue}{(1+x)(1+x)(1+x)\cdots(1+x)}} \leqslant \frac{\color{red}{(1+1+1+\cdots+1)}+\color{blue}{(1+x)+(1+x)+\cdots (1+x)}}{n}$$ which is AM-GM as all terms are non-negative. Note the blue terms on each side are $m$ in number and the red terms count to $n-m$.

Answer 2

PRIMER:

It is easy to show that the sequence $e_n(x)=\left(1+\frac xn\right)^n$ increases monotonically for $x>-1$. To show this we simply analyze the ratio

$$\begin{align} \frac{e_{n+1}(x)}{e_n(x)}&=\frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^n}\\\\ &=\left(1+\frac{-x}{(n+x)(n+1)}\right)^{n+1}\left(1+\frac xn\right) \tag 1\\\\ &\ge \left(1+\frac{-x}{n+x}\right)\left(1+\frac xn\right)\tag 2\\\\ &=1 \end{align}$$

where in going from $(1)$ to $(2)$ we used Bernoulli's Inequality. Note that $(2)$ is valid whenever $n>-x$ or $x>-n$.

Equipped with the monotonicity of $e_n$ we now proceed.

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Let $n$ and $m$ be positive integers such that $p=m/n\le 1$. Then, using $e_m(y)\le e_n(y)$ for $y>-1$ and letting $y=mx$ we see that for $x\ge0$

$$\left(1+x\right)^m\le \left(1+\frac{mx}{n}\right)^n\tag3$$

Taking the $n$th root of both sides of $(3)$ yields

$$(1+x)^p\le 1+px$$

for $0\le p\le 1$, as was to be shown.

Answer 3

For x positive the sequence $(1+\frac{x}{n})^n$ increases. So we have $(1+\frac{xp}{q})^q$ $\lt $ $(1+\frac{xp}{p})^p$ $=(1+x)^p$ $\Rightarrow $ $1+\frac{xp}{q}$ $\lt$ $(1+x)^{p/q}$ $\Rightarrow $ $1+xr\lt(1+x)^r $.