Lower bound for Ramsey numbers: $R( n + 2 , 3 )>3n$ I need to prove the following inequality: $$R( n + 2 , 3 )>3n$$ where $n>1$ and $R(s,t)$ is a Ramsey number.

The most general way to prove such inequalities is to paint a graph with $3n$ nodes then show it could not have a sub graph with $n+2$ in red or $3$ in blue, I've also used induction but no answers yet.
 A: Sorry I dont know how to make a picture.
Here I hope is an example Take $x_1,\cdots, x_n, y_1,\cdots, y_n, z_1,\cdots, z_n$
Make the $x$'s red. As are also the $y$'s and the $z$'s red. 
Make $(y_i, z_i)$ blue and $(y_i, z_j)$ red. 
Make $(y_i, x_i)$ blue and $(y_i, x_j)$ red.
Then 
$(z_i, x_i)$ red and $(z_i, x_j)$ blue.
I submit this much before the question is closed.
Basically you have a set $X$ of size $n$ which is all red. Now for each $x\in X$ there is a blue pair $(y,z)$ with $z$ blue only to $x$ and $y$ red only to $x$. 
Lets say there is a blue triangle. It has at most one $x$ one $y$ and one $z$.
They must be $y_i$ and $z_i$ for the same $i$ since all other pairs are red.
Now $x$ cannot be $x_i$ since $y_i$ is red with $x_i$.
Finally there is no $n+1$ red. It cannot have both $y_i$ and $z_i$ since these are blue. Thus from the $y$'s and the $z$'s there are at most $n$. And no pairs. This means there are at least two $x$'s Say $x_1$ and $x_2$ for convienience. Then I cannot have $z_1$ or $z_2$ since they are blue with $x_1$ and $x_2$ respectively. Thus I have to have $$y_1$ and $y_2$; but $y_2$ and $x_1$ are blue a contradiction.  
