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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 $\log$?

hamid kamali
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Pieter Verschaffelt
  • 481

2 Answers2

4

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 .

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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.

mfl
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