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Suppose $a,b>0$ are integers. Do we then have that $$ \ln(a+b)\leq \ln(a)+\ln(b)? $$

I think, since this is equivalent to $a+b\leq ab\Leftrightarrow 1\leq b-\frac{b}{a}$

it holds for $1<b<a$ only and, moroever, it is a strict inequality for these values.

Salamo
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3 Answers3

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You are correct. This indeed fails if $a=1$ and $b\in\mathbb{N}$, as $\log(1)=0$, so the inequality becomes $$\log(1+b)\leq\log(b)$$ which is clearly false. However, if we strengthen the premise to $a,b>1$, then this inequality holds.

Your strengthening to $1<b<a$ omits the equality case $a=b=2$, as $$\log(2+2) = \log(2\cdot 2) = \log(2)+\log(2)$$

Isaac Browne
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Suppose $a$ and $b$ are positive integers. Then:

\begin{align} \ln(a+b)\leq \ln(a)+\ln(b) \\ \\ (*)\quad \iff a+b \leq ab \quad\\ \\ \iff 1 \leq b - \frac{b}{a} = b\left(1-\frac{1}{a}\right),\\ \\ \iff b \geq 2\ \text{ and }\ a \geq 2. \end{align}

The $\impliedby$ at $(*)$ is justified by the fact that $\ln$ is an increasing function. The $\implies$ at $(*)$ is justified by the fact that $\exp$ is an increasing function.

Adam Rubinson
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$$a+b\le ab\iff 1\le(a-1)(b-1),$$ which is wrong for $a=1$ or $b=1$ and works for other naturals.