Recurrence Relation Inequality claim in "Polynomomial Methods in Combinatorics" While this question is in the context of combinatorics, I do not believe it plays any special role beyond producing the problem.
In the proof of Theorem 10.18 in Larry Guth's book "Polynomial Methods in Combinatorics" the following assertions are made. (page 134 in particular)
We have the recurrence relation:
$$|S_{j+1}| \leq \frac{1}{100} |S_j| + \frac{1}{200} K L^{3/2 + \varepsilon}r^{-2}$$
where $L$ is a positive integer, and $K$ is a large constant.
Now Guth claims that if we define $J = 1000 \log L$ and notice that $|S_1| \leq L^2$ we can conclude that
$$|S_J| \leq \frac{1}{100} KL^{3/2 + \varepsilon} r^{-2}.$$
We also have the fact that the sequence of $|S_j|$ is non-increasing and that $2 < r \leq 2L^{1/2}$. I'm not seeing how one can make this leap, but perhaps I am missing something straightforward.
 A: (Below I am assuming $\log$ means $\log$ base $e$.)

By induction we obtain that $|S_k|\leqslant\frac{1}{100^k}|S_1|+\frac{1}{200}KL^{3/2+\varepsilon}r^{-2}\left[\frac{1}{100^{k-1}}+\dots+\frac{1}{100}+1\right]$ for every $k\geqslant 1$. Using the formula for geometric series and the bound $|S_1|\leqslant L^2$, we thus get $$|S_k|\leqslant\frac{L^2}{100^k}+\frac{100^{k}-1}{99\times 200\times 100^{k-1}}KL^{3/2+\varepsilon}r^{-2}.$$ Note that $L^2/100^J=L^2/(100^{\log_{100}(L)})^{1000/\log_{100}(e)}=L^{2-1000/\log_{100}(e)}$. Moreover, note that $$\frac{100^k-1}{99\times 200\times 100^{k-1}}\leqslant\frac{100}{99\times 200}=\frac{1}{2\times 99}$$ for any $k\geqslant 1$. Hence $|S_J|\leqslant \frac{1}{L^{-2+1000/\log_{100}(e)}}+\frac{1}{2\times 99}KL^{3/2+\varepsilon}r^{-2}$, and so all we need to show is $$\frac{1}{L^{-2+1000/\log_{100}(e)}}\leqslant\frac{49}{9900}KL^{3/2+\varepsilon}r^{-2}.$$ But $r^{-2}\geqslant 1/4L$, so the right hand side is at least $\frac{49}{4\times 9900}KL^{1/2+\varepsilon}$. Since $K>1$ (say) and $L>2$ (say), the result follows.
