How to prove the elementary inequality? The inequality is the following:
$$\frac{(1+x)^q-1}{x+x^q} \leq C(q),$$
where $q\in [1,+\infty)$  and $x > 0$, and the constant $C$ depends only on $q$. It's very nice if someone can provide the minimal value of $C(q)$, I guess the minimal value is $q$.
 A: We will  show that for fixed $q$ the function $\frac{(1+x)^q-1}{x^q+x}$ is bounded. By L'Hospitals Rule, we have 
$$\lim_{x\to 0^+}\frac{(1+x)^q-1}{x^q+x}=\lim_{x\to 0^+} \frac{q(1+x)^{q-1}}{qx^{q-1}+1}=q.$$
And it is clear that
$$\lim_{x\to \infty}\frac{(1+x)^q-1}{x^q+x}=1$$
if $q\gt 1$. (The case $q=1$ is easy to deal with.)
Boundedness follows.
A: This is not an answer but it is too long for a comment.
For the function $$F(x,q) = \frac{(1 + x)^q - 1}{x^q + x}$$ André Nicolas showed that $$F(0,q)=q$$ $$F(\infty,q)=1^{+}$$ What I looked at is the problem of the derivative and here, there are a few things which are interesting as soon as $q\gt 2$.
The derivative  $$F'(0,q)=\frac 12 q(q-1)\gt 1$$ means that the function goes through a maximum value. On the other hand $$F'(1,q)=\frac{1}{4} \left(q-2^q+1\right)<0$$ means that the minimum is between $0$ and $1$. So, the value of the function at the maximum is always larger than $$F(1,q)=\frac{1}{2} \left(2^q-1\right)$$ Expanding the function as a Taylor series around $x=1$ shows that the derivative cancels close to $$x^*=\frac{4+8 q+2^q (q^2-3q-4) }{2+6q+2^q \left(q^2-3 q-2\right)}$$ which clearly shows the asymptotic behavior of the location of the maximum.
What it seems (I have not been able to prove it) is that the maximum value of the function is something like $$F(x^*,q)\approx 2^{q-1}(1+\epsilon)$$ and more precisly, $$F(x^*,q)\approx \frac{1}{2} \left(2^q-1+\frac{\left(2^q-q-1\right)^2}{2+6 q+2^q (q^2-3 q-2)}\right)$$
All the above was checked numerically. For example, 


*

*for $q=10$, the maximum is located at $0.972364$ while the approximation gives $\approx 0.970930$; the function value at the maximum is $515.050$ while the approximation gives $518.862$ 

*for $q=30$, the maximum is located at $0.997536$ while the approximation gives $\approx 0.997525$; the function value at the maximum is $5.37202\times 10^8$ while the approximation gives $5.37535\times 10^8$ 


Interesting problem, indeed !
A: André Nicolas showed that $C(q)$ is finite and $\ge q$.
Claude Leibovici showed that $C(q)\ge \frac12(2^{q}-1)$ and conjectured that $C(q)\approx 2^{q-1}(1+\epsilon)$.
We are still missing an explicit bound on $C(q)$ from above; here's a very simple one:
By the extended mean value theorem with $f(x)=(1+x)^q$ and $g(x)=x+x^q$ we find that
$$ F(x,q)=\frac{f(x)-f(0)}{g(x)-g(0)}=\frac{f'(\xi)}{g'(\xi)}=\frac{q(1+\xi)^{q-1}}{1+q\xi^{q-1}}$$
with $0<\xi<x$. 
The supremum of the function $G(x,q)=\frac{q(1+x)^{q-1}}{1+qx^{q-1}}$ is easier to handle as 
$$\begin{align}G'(x,q)&=\frac{q(q-1)(1+x)^{q-2}\cdot (1+qx^{q-1})-q(1+x)^{q-1}\cdot q(q-1)x^{q-2}}{(1+qx^{q-1})^2}\\
&=\frac{q(q-1)(1+x)^{q-2}}{(1+qx^{q-1})^2}\cdot(1+qx^{q-2}-(1+x)qx^{q-2})\\
&=\frac{q(q-1)(1+x)^{q-2}}{(1+qx^{q-1})^2}\cdot(1-qx^{q-1}) \end{align}$$
Hence for $q>1$  (note that $G(x,1)=1$) $G'$ has a unique sign change and so $G$ a maximum at $x^*=\sqrt[q-1]{1/q} <1$. Therefore
$$F(x,q)\le G(x^*,q)=\frac q2 (1+x^*)^{q-1}<q2^{q-1}$$
(Asymptotically, this can be improved to $G(x^*,q)\sim \sqrt q\,2^{q-1}$, which is still larger than the conjectured value, therefore I don't go into that detail)
