T(3) = T(2) + 2*T(1) + n. It uses previous base cases therefore we don't need any more base cases.

Base cases:

n=1 : T(1) = 1 $\geq$ c*1*log1 for any c

n=2 : T(2) = 1 $\geq$ c*2*log2 for c $\leq$ 1/2

Step: We show that the inequality holds for some n $\geq$ $n_0$

IH: Assume that T(k) $\geq$ c*k*log k

T(n) = T($\lceil$n/2$\rceil$) + 2*T($\lceil$n/4$\rceil$) + n    (IH on $\lceil$n/2$\rceil$ and $\lceil$n/4$\rceil$)

T(n) $\geq$ c*$\lceil$n/2$\rceil$*log$\lceil$n/2$\rceil$ + 2*c*$\lceil$n/4$\rceil$*log$\lceil$n/4$\rceil$ + n

n/2 $\leq$ $\lceil$n/2$\rceil$ and n/4 $\leq$ $\lceil$n/4$\rceil$ and n/4 < n/2

c*n/2*log(n/4) + 2*c*n/4*log(n/4) + n

c*(n/2+2*n/2)*log(n/4) + n

c*n*log(n/4) + n

c*n*logn -2*c*n + n

c*n*logn -n(2*c - 1) $\geq$ c*n*logn

-2c+1 $\geq$ 0

2c $\geq$ -1

c$\leq$ 1/2