Solving a particular inequation I don't know the method to solve the following inequation
$$C_1 (1+ \sqrt{a})^{n-1} + C_2(1-\sqrt{a})^{n-1} \geq k$$
where $C_1$, $C_2$, and $a$ are constants. I would like to express a lower bound for $n$ as a function of $k$.
 A: This is not simple.  The addition doesn't play nicely with the powers so you will need to find a numeric solution.  The best approach will depend upon the constants you have.  We set $m=n-1$ to save typing.  
The easiest case is when $k$ is rather large, especially where $C_2$ is not much greater than $C_1$.  In that case the first term will dominate.  You can ignore the second to get a starting value, then iterate.  Set $$C_1(1+\sqrt a)^{m_0}=k
\\m_0=\frac {\log k - \log C_1}{\log (1+\sqrt a)}\\m_{i+1}=\frac {\log (k-C_2(1-\sqrt a)^{m_i}) - \log C_1}{\log (1+\sqrt a)}$$ and iterate to convergence  
Another easy case is when $a$ is small and $m$ is not too large.  You can use the binomial expansion then and write $$(1+\sqrt a)^m \approx 1+m\sqrt a\\C_1 (1+ \sqrt{a})^{m} + C_2(1-\sqrt{a})^{m}\approx C_1+C_2 +(C_1-C_2)m\sqrt a$$
and you have a linear equation to solve for $m$.  Again, you can use that as a starting value for iteration if it is not accurate enough.  The test for accuracy is $m \sqrt a \ll 1$  
If $C_2$ is rather large compared with $C_1$ you can have two ranges of $m$.  An example is in the plot below where I took $C_1=1, C_2=100, \sqrt a=0.1$  You can see that for $k=40$ you have two regions of solution.  The minimum value for the higher region will be found easily by the first approach.  The root around $10$ will be harder to find.

