Bernoulli's inequality for rational exponents 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?
 A: 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 $n-m$.
A: 
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.

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.
A: 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 $. 
