Is it obvious that (a+c)/(b+d) is between a/b and c/d?

No it isn’t obvious, because one does not know the magnitude of the ratios, and also one does not know the relationship between the numerators and denominators.

let’s put some numbers in place
a=2
b=1
c=6
d=3
so (a+c)/(b+d) = (2+6)/(1+3)=(8)/(4) = 2

a/b=2/1=2
c/d=6/3=2

so I guess if you say 2 falls between 2 and 2 it holds true.

I should have been more explicit. By “between” I meant less than or equal to the higher number and greater than or equal to the smaller. In equation form: For a, b, c and d all greater than 0, we can assume without loss of generality (otherwise swap and b for c and d) that a/b <= c/d. Then a/b <= (a+c)/(b+d) <= c/d.

If a team scores 2 runs per game at the start of the season and scores 2 runs per game for the second part of the season then the average for the year will also be 2 runs per game.

No, it’s not obvious but it’s eventually apparent.

a/b > c/d
ad > bc
ad + ab > bc + ab
a(b + d) > b(a + c)
a/b > (a + c)/(b + d)

a/b > c/d
ad > bc
ad + dc > bc + dc
d(a + c) > c(b + d)
(a + c)/(b + d) > c/d

That is the standard proof, but here is another that I think is more intuitive.

(a+c)/(b+d) = a/b *(b/(b+d)) + c/d(d/(b+d))
Since b/(b+d) + d/(b+d) = 1, what we have is a weighted average of a/b and c/d. It is easy to show that the weighted average of two numbers lies between them.