An $n$-vertex graph of minimum degree $\delta >1$ has a dominating set of size at most $n(\log (\delta +1)+1)\mathrm{/}(\delta +1)$

Put $p=\log (\delta +1)\mathrm{/}(\delta +1)$. Since $\delta +1>1$ and the logarithm is increasing with $\log 1=0$, one has $p>0$. Applying [L4] at $-p$ is not needed here: applying its first inequality at $\log (\delta +1)$ gives $1+\log (\delta +1)\le \delta +1$, hence $p\le \delta \mathrm{/}(\delta +1)<1$.

Select every vertex independently with probability $p$, obtaining $S$, and let $U$ be the vertices neither in $S$ nor adjacent to a member of $S$. Then $D=S\cup U$ is dominating.

A fixed vertex belongs to $U$ only if none of at least $\delta +1$ vertices in its closed neighbourhood is selected, so $\mathbb{P}(v\in U)\le (1-p{)}^{\delta +1}\le \exp (-p(\delta +1))=1\mathrm{/}(\delta +1)$.

By linearity, $\mathbb{E}[\mathrm{\mid }D\mathrm{\mid }]\le np+n\mathrm{/}(\delta +1)=n(\log (\delta +1)+1)\mathrm{/}(\delta +1)$.

Some outcome has $\mathrm{\mid }D\mathrm{\mid }$ at most this expectation, and its $D$ is a dominating set by step 1.2.