show that [x+y] is greater than or equal to [x]+[y] for all real numbers x and y

Question

not sure what to do with these... the problem in my book reads show that [x+y] ≥ [x]+[y] for all reals x and y.....(brackets, not abs. value bars)

Answer 1

if x and y are both positive integers, then [x+y] = [x] + [y] 
if x is a positive integer and y is a negative integer, then [x+y] = x+y, [x] + [y] = x + y - 1, therefore [x+y] > [x]+[y]

if x and y are both negative integers, then [x+y] = x+y -1, [x] + [y] = x-1 + y-1 = x+y - 2, therefore [x+y] is >

if x = n+a and y = m + b, where n and m are integers and a and b are fractions 0 < a < 1, 0 < b < 1
then if a + b = c > 1, [x+y] = x+y+c, [x] + [y] = x+y, therefore [x+y] > [x] + [y]

if a + b < 1, then they are equal

similar statements can be made for negative values of a and b, but regardless [x+y] is always equal to or greater than [x] + [y]

I imagine there is a more elegant way to show this statement is true. Perhaps an actual proof exists.