Solution
= Solution
For every real $r\ge0$,
$$
\log(1+r)=\int_0^r\frac{dt}{1+t}\le r.
$$
Taking $r=|w|$ proves the required <logarithm inequality>, even without the lower bound $|w|\ge1/2$. Hence \b[$C_2=1$ works].
= Solution
For every real $r\ge0$,
$$
\log(1+r)=\int_0^r\frac{dt}{1+t}\le r.
$$
Taking $r=|w|$ proves the required <logarithm inequality>, even without the lower bound $|w|\ge1/2$. Hence \b[$C_2=1$ works].