A stronger form of Bernoulli's inequality Let $x\geq 0$ be fixed. By Bernoulli's inequality we know that for all $n \in \mathbb{N}_0$,
$$
(1+x)^n \geq 1+nx
$$
Now, let $\alpha \geq 0$ be fixed as well. By a limit argument, we know that
$$
(1+x)^n \geq 1+\alpha nx
$$
for large $n$. But how large does $n$ have to be? More specifically, can we find some expression for $N \in \mathbb{N}$ depending on $x$ and $\alpha$ such that the inequality holds for all $n \geq N$?
If it helps, $N$ does not necessarily have to be optimal. I am specifically interested in the case where $\alpha \in \mathbb{N}$, and even more specifically in the case $\alpha=2$.

Well, of course. 
Equations in $x$ of the form $p^x=ax+b$ have solutions given by the Lambert-$W$ function.
In your specific case, we have
$$(1+x)^n=(\alpha x)n+1$$
And the corresponding solution in $n^*$ is
$$n^*=-\frac{W_{-1}\left(-\frac{\ln(1+x)}{\alpha x}\,(1+x)^{-\frac1{\alpha x}}\right)}{\ln(1+x)}-\frac1{\alpha x}$$
Hence, for all $n\geq n^*$ we have $(1+x)^n\geq 1+n\alpha x$.

There's the bound
$$W_{-1}(-e^{-u-1})\geq -1-\sqrt{2u}-u.$$
Setting $-e^{-u-1}=-\frac{\ln(1+x)}{\alpha x}\,(1+x)^{-\frac1{\alpha x}}$, we get
\begin{align}
u&=-1-\ln\left(\frac{\ln(1+x)}{\alpha x}\,(1+x)^{-\frac1{\alpha x}}\right)\\
&=-1-\left(\ln\left(\frac{\ln(1+x)}{\alpha x}\right)+\ln\left((1+x)^{-\frac1{\alpha x}}\right)\right)\\
&=-1-\left(\ln(\ln(1+x))-\ln(\alpha x)-\frac1{\alpha x}\ln(1+x)\right)\\
&=\ln(\alpha x)+\frac{\ln(1+x)}{\alpha x}-1-\ln\Big(\ln(1+x)\Big).
\end{align}
So, depending on how precise you wish to be, you may use
$$n^*\geq \frac{1+\sqrt{2u}+u}{\ln(1+x)}-\frac1{\alpha x},$$
where $u$ is as above.
