There exists a constant $C$ s.t. if $G$ is an $n$-vertex graph with no subgraph isomorphic to $K_4$, then $\chi(G) \leq C \times n^{2/3}$ If $G$ has no subgraph isomorphic to $K_4$, then it does not have a sugraph isomorphic to $K_l \text{ s.t. }l\geq 4$.
Facts I know: $\chi(G) \geq \omega(G)$, where $\omega(G)$ is the size of the largest clique in $G$.
Also, $\chi(G) \geq \frac{|V|}{\alpha(G)}$, where $\alpha(G)$ is the size of the max independent set.
Also, Brooks theorem says that $\chi(G) \leq \Delta(G)$ if $G$ is not an odd cycle or a complete graph. Otherwise $\chi(G) \leq \Delta(G)+1$. (Where $\Delta(G)$ is the max degree of a vertex in $G$).
I think these are all the facts related to $\chi(G)$ that should be needed to prove the statement in the title, but I can't figure out how to approach this proof. Advice on how to proceed would be appreciated!
 A: An upper bound on the chromatic number of $K_4$-free graphs is essentially equivalent to an lower bound on their independence number. An upper bound of $\chi(G) \le C n^{2/3}$ is equivalent to a lower bound of $\alpha(G) \ge c n^{1/3}$, which follows from the Ramsey result $R(4,k) \le \binom{k+2}{3}$.
One direction of the equivalence between these comes from result you've cited: $\chi(G) \ge \frac{n}{\alpha(G)}$, so if $\chi(G) \le C n^{2/3}$, then $\frac{n}{\alpha(G)} \le C n^{2/3}$, and therefore $\alpha(G) \ge C^{-1} n^{1/3}$. 
The other direction of the equivalence, though, is the one you'll actually need to use here. If all $K_4$-free graphs have an independent set of size $cn^{1/3}$, then you can color a $K_4$-free graph with $O(n^{2/3})$ colors (something like $\frac {10}c n^{2/3}$ colors is definitely possible) by an iterative argument: pick off an independent set of size $c n^{1/3}$, give it a color, and repeat on the rest of the graph (which is also $K_4$-free).
A not-terribly-sharp bound on how many colors you use this way can be obtained by combining the following arguments:


*

*It takes $O(n^{2/3})$ steps to go from $n$ vertices to fewer than $n/2$.

*The same is true to go from $n/2$ to $n/4$ vertices, and $n/4$ to $n/8$, and so on, but with a factor of $(1/2)^{2/3}$ being tacked on each time, so the total number of steps is $$1 + \frac1{2^{2/3}} + \frac1{2^{4/3}} + \dots$$ times the first $O(n^{2/3})$, which only gives a constant factor.

*For a stopping condition, once you get down to fewer than $n^{2/3}$ vertices, you can give each remaining vertex its own color.


You can see a similar argument for $K_3$-free graphs being carried out in this answer.
