Prove the inequality $(1 + x)^{1/\pi} < 1 + \frac{x}{\pi}$ for $x > 0$ 
Prove the inequality $(1 + x)^{1/\pi} < 1 + \frac{x}{\pi}$ for $x > 0$

The inequality looks very straightforward to prove; however, I have been struggling for a while. The first thing that came to my mind was the Binomial Series, but there are both positive and negative terms in there, so it is not clear that the result follows.
I also tried defining a function $f(x)$ and taking derivatives, but that's not valid since the power rule only holds for rationals and $1/\pi$ is irrational.
EDIT: this also looks a lot like Bernoulli's inequality; however, I think the exponent has to be an integer in that.

Let $f(x) = \frac{1}{\pi}\log(1 + x) - \log(1 + \frac{x}{\pi})$, and note that $f(0) = 0$. So, by showing that $f$ is decreasing, our result will follow. We have
$$f'(x) = \frac{1}{\pi} \frac{1}{1 + x} -\frac{1}{1 + \frac{x}{\pi}} \cdot \frac{1}{\pi} $$
$$ \frac{1}{\pi}\left(\frac{1}{1 + x} - \frac{1}{1 + \frac{x}{\pi}}\right) < 0 $$
$$\Longleftrightarrow \frac{1}{1 + x} < \frac{1}{1 + \frac{x}{\pi}} $$
$$ \Longleftrightarrow 1 + x > 1 + x/\pi $$
$$x > x/\pi, $$
which is clearly true since $x > 0$.
So, since we have shown that the logarithm of the function is negative for $x > 0$, it follows that the function itself is decreasing.
 A: Defining $$f(x)=\frac{1}{\pi}\ln(1+x)-\ln\left(1+\frac{x}{\pi}\right)$$ and now use calculus. For $$x>0$$ we get $$x\left(1-\frac{1}{\pi}\right)>0$$ which is true.
$$x=0$$ gives the maximum of our function. The first derivative of our $$f(x)$$ is given by
$$\frac{1}{\pi(1+x)}-\frac{1}{1+\frac{x}{\pi}}\cdot \frac{1}{\pi}$$
solving $$f'(x)=0$$ we get $$1+\frac{x}{\pi}-1-x=0$$ and $$\frac{x}{\pi}-x=0$$ is $$x\left(\frac{1}{\pi}-1\right)=0$$
A: Answer to the original question: $(1 + x)^{1/\pi} > 1 + \frac{x}{\pi}, \, x>0$
The inequality is not true, indeed for $x=1$ we have
$$(1 + 1)^{1/\pi}\approx 1.247$$
$$1 + \frac{1}{\pi}\approx 1.318$$
indeed for $0\le r\le 1$ the following Bernoulli's inequality holds as $x\ge -1$
$$(1+x)^r \le 1+rx$$
Answer after editing: $(1 + x)^{1/\pi} < 1 + \frac{x}{\pi}, \, x>0$
For $0\le r\le 1$ the following Bernoulli's inequality holds as $x\ge -1$
$$(1+x)^r \le 1+rx$$
and equality holds only for $r=0,1$ or $x=0$.
A: The inequality is not true, because for $0<r<1$ we have that
$$(1+x)^r<1+rx$$
A: Power rule is valid for all real powers. 
Define $f(x)=(1+x)^p-1-px;\ 0<p<1, x\ge0$
$\displaystyle f(0)=0, f'(x)=p(1+x)^{p-1}-p=\frac p{(1+x)^{1-p}}-p\le0$
Therefore, $f(x)\le0,\ \forall x\ge0$
