I have to prove the following, where $x\geq 1$, $0 < \epsilon< 1$ and $q\in \mathbb{N}$
$$(x + \epsilon)^q \leq x^{q}(1 + 2^{q}\epsilon)$$
My attempt:
We have
$$(x+\epsilon)^q = x^q + \binom{q}{1}x^{q-1}\epsilon + \binom{q}{2}x^{q-2}\epsilon^2 \cdots + \binom{q}{q-1}x\epsilon^{q-1} +\epsilon^q $$
And as $\epsilon <1$ and $q\geq 1$, then $\epsilon^q \leq \epsilon$. Therefore,
\begin{align} (x+\epsilon)^q \leq x^q + \binom{q}{1}x^{q-1}\epsilon + \binom{q}{2}x^{q-2}\epsilon +\cdots + \binom{q}{q-1}x\epsilon +\epsilon \end{align}
In other words,
$$(x+\epsilon)^q \leq x^q +\epsilon\biggl(1+ \binom{q}{1}x^{q-1} + \binom{q}{2}x^{q-2} +\cdots + \binom{q}{q-1}x\biggr)$$
I know that I need to get the sum of the binomial coefficients to get $2^q$ but I can't find a way to deal with the $x$ terms attached to them. Any help would be greatly appreciated. Thanks
Hint
Factorizing the RHS of your last expression by $x^q$ yields $$x^q +\epsilon\biggl(1+ \binom{q}{1}x^{q-1} + \binom{q}{2}x^{q-2} +\cdots + \binom{q}{q-1}x\biggr)=x^q\left[1+\varepsilon \left(\frac{1}{x^q}+\binom{q}{1}\frac{1}{x}+...+\binom{q}{q-1}\frac{1}{x^{q-1}}\right)\right].$$
Using the fact that $\frac{1}{x^p}\leq 1$ for all $p\in\mathbb N$ allows you to conclude.