Proof $a + b \le c $ implies $\log(a) + \log(b) \le 2\log(c) - 2$ I have to proof the following statement:

Assume $a, b$ and $c$ are natural numbers that are different of $0$. 
      If $a + b \le c $, then $\log(a) + \log(b) \le 2\log(c) - 2$. 
      All $\log$ functions are the second $\log$ functions, thus $\log2$.

I created the following proof but I'm not certain if it's completely correct:
Because $a > 0$ and $b > 0$, we find:
$a < c$ and $b < c$
Thus:
 $$\log(a) < \log(c) \Rightarrow  \log(a) \le \log(c) - 1$$
and
$$\log(b) < \log(c) \Rightarrow  \log(ab\le \log(c) - 1$$
 When we add these $2$ expressions, this gives us: $$\log(a) + \log(b) \le 2\log(c) - 2$$
QED
I wonder if the step $\log(a) < \log(c) \Rightarrow  \log(a) \le \log(c) - 1$ is correct? I know it is correct for $a < b$ if and only if $a \le b - 1$, but is does this also correspond to $\log$?

Hint
$$\log_2(ab)=\log_2 a+\log_2 b\le 2log_2 c-2=\log_2c^2-\log_2 4=\log_2 \frac{c^2}{4}.$$
Thus, if you show that $$ab\le \frac{c^2}{4}$$ you are done.
