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?
Best Answer
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 $n-m$.