Median of Medians Recurrence Relation for 3-grouping So I am trying to figure out the recurrence relation for the median of medians algorithm using groups of 3 instead of groups of 5. Per CLRS's method, my recurrence relation looks like 

$$
T(n) = T(\lceil \frac{n}{3} \rceil) + T(\frac{2n}{3} + 4) + c_1n
$$

And I want to show $T(n) = O(n \lg n)$

Via substition, I assume inductively that 

$T(n') = cn' \lg n'$

So by substition

$$T(n') \leq c\lceil \frac{n'}{3} \rceil \lg \lceil \frac{n'}{3} \rceil + c(\frac{2n'}{3} + 4) \lg (\frac{2n'}{3} + 4) + c_1n' $$

$$ \leq c (\frac{n'}{3} + 1) \lg (\frac{n'}{3} + 1) + c(\frac{2n'}{3} + 4) \lg (\frac{2n'}{3} + 4) + c_1n'$$

aaand now im stuck. Is there anyway i can drop the +1 and +4 from this expression to make it simpler?
You can use the inequality $\log (1+x) &lt; x$ (valid for $x &gt; 0$, where $\log$ is the natural logarithm), which follows from a Taylor expansion, to reason as follows:
$$
\log \left(\frac{n'}{3} + 1\right) =
\log \left(\frac{n'}{3} \cdot \left(1 + \frac{3}{n'}\right)\right) = \log \frac{n'}{3} + \log \left(1 + \frac{3}{n'}\right) &lt; \log \frac{n'}{3} + \frac{3}{n'}.
$$
This shows (after some calculation) that you can remove the $+1$ and $+4$ in return for an additional constant additive term, that is, you can estimate your final line by
$$
c \left(\frac{n'}{3}+1\right) \lg \frac{n'}{3} + c\left(\frac{2n'}{3}+4\right) \lg \frac{2n'}{3} + M + c_1 n',
$$
where $M&gt;0$ is some constant.