If a,b > 0, c >= a+b, 
then log a + log b <= 2 log c - 2.

Proof:
(a-b)^2 >= 0     [a square is always nonnegative]
(a-b)^2 + 4ab >= 4ab      [add 4ab to both sides]
a^2-2ab+b^2 + 4ab = a^2+2ab+b^2 = (a+b)^2 >= 4ab    [expanding, factoring]
Aside: (a+b)^2 / 4 >= ab => (a+b)/2 >= sqrt(ab)  is the AM-GM inequality:
the arithmetic mean is at least the geometric mean.
(a+b)^2 >= 4ab 
log((a+b)^2) >= log(4ab)		[log is monotonic increasing]
2 log(a+b) >= log 4 + log a + log b	[various log properties]
2 log c >= 2 log(a+b) >= 2 + log a + log b	[c >= a+b]
log a + log b <= 2 log c - 2	[rearranging]