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 $\log$?

Start from $c \geq a+b$ so :
$$2\log_2 c -2 \geq 2 \log_2(a+b)-2$$
If you can show that :$$2 \log_2(a+b)-2 \geq \log_2 a +\log_2 b$$ then you're done .
This is equivalent with :
$$2^{2 \log_2(a+b)-2} \geq 2^{\log_2 a +\log_2 b}$$ or :
$$\frac{1}{4} (a+b)^2 \geq ab$$ (note that I used the obvious fact that  $2^{\log x} =x$ )
But this last inequality is equivalent with : $$(a-b)^2 \geq 0$$ which is true .
