Inequality for sum to the exponent of $q>1$

Question

In this question: How to use $\lesssim$ when using exponents I already asked about an inequality for a sum to the exponent of $2$.

Now I wonder if $\forall q>1$, there exists a constant $C>0$ such that for all $x\in\mathbb{R}_+$ it holds that \begin{align*} (1+x)^q\leq C (1+x^q). \end{align*}

I tried to work with an integer $k\geq q$ and exploit convexity in the positive real numbers, but that was the wrong way since it only lead me to $(1+x)^q\leq C(1+x^k)$. In some way one might use Young's inequality but I didn't manage to do it.

Answer 1

Just as a short summary, thanks to @leoli:

\begin{align*} \frac{1}{2^{q}}(1+x)^q=(\frac{1}{2}+\frac{x}{2})^q\leq\frac{1}{2}\cdot1^q+\frac{1}{2}x^q, \end{align*}

so $C=2^{q-1}$.