Comparison of the numbers of colored squares in a checkerboard colored with red green and blue The squares of a checkerboard $n \times n$ are colored alternatively in red, blue and green so that: next to a red square, there is a blue square; next to a blue square, there is a green square; next to a green square, there is a red square. Denoting $r, b, g$ the numbers of red, blue and green squares respectively, show that: 
$\bullet\ r \leq 3b, b \leq 3g, g \leq 3r$;
$\bullet\ r \leq b + 4g, b \leq g + 4r, g \leq r + 4b$.
The first point is rather simple to prove. For example, for the first inequality, we can have:
three red squares around a blue square folowed by a green square
so potentially there are three times more red squares than blue squares.
But I do not see how to prove the second point ($r \leq b + 4g, b \leq g + 4r, g \leq r + 4b$).
 A: I suppose that for a square to be "next to" another square means the two squares share an entire edge, not just that they touch at a corner.
To formalize your first conclusion a bit,
consider a playing piece that can move one space up, down, left, or right,
but not diagonal, and can only move from green to blue or from blue to red.
Then the only red squares that can exist are one move away from a blue square.
For any board, we can make partition the squares into disjoint subsets as follows:
each subset contains exactly one blue square,
and each red square is in a partition that also contains at least one blue square that the red square is one move away from.
But from any blue square we can only move to an adjacent square, and since there are only four such squares and one must be green, that leaves at most there adjacent red squares that could be one move away from the blue square. 
Your figure illustrates this fact.
Since at most three red squares can be in the same subset as any blue square,
$r$ is no greater than three times the number of subsets,
and since $b$ is the number of subsets, $r \leq 3b.$
For the second part of the claim, make a new partition in which each subset contains exactly one green square, each blue square in the subset must be exactly one move away from the green square, and each red square in the subset must be exactly one move away from one of the blue squares.
The number of blue squares in the subset must be an integer no less than $0$ and no greater than $3$; up to symmetry, this leaves only one case each for how $0,$ $1,$ or $3$ blue squares can be arranged around the green square, and two cases for $2$ blue squares. In each case you can show that in the $k$th subset, if it contains $r_k$ red, $b_k$ blue, and $g_k$ green squares, then $r_k \leq b_k + 4 g_k.$
A: Yes, a square is "next to" another square means the two squares share an entire edge, not just that they touch at a corner. By applying your method for the second question, the results I got are summerized below.
Green    Blue                  Red              Configuration
1        0                     0                C0
1        1                     Between 0 and 3  C1
1        2 (2 configurations)  Between 0 and 6  C2 and C2'
1        3                     Between 0 and 7  C3

C0, C1, C2, C2' and C3 
But I don't understand why the subsets of this partition should be disjoint. In particular, it seems to me that the red squares which are two move away from the only one green square of the subset, can also be two move away from another green square.
