We need to prove that, for all $s\in \mathbb{R}$ with $s\ne \pm 1$,
$$\frac{1}{2}(s-1)^2 \ln^2\left(1-\frac{2s}{1+s^2}\right) + \frac{1}{2}(s+1)^2 \ln^2(1+\frac{2s}{1+s^2})\le 4s^2. \tag{1}$$
It suffices to prove that, for all $s\in \mathbb{R}$ with $s \ne \pm 1$,
$$\frac{1}{2}(s+1)^2 \ln^2\left(1+\frac{2s}{1+s^2}\right)\le 2s^2. \tag{2}$$
(Note: If (2) is true, letting $s \to -s$, we have $\frac{1}{2}(s - 1)^2 \ln^2\left(1 - \frac{2s}{1+s^2}\right)\le 2s^2$ which, together with (2), yields (1).)
- If $s \ge 0$ with $s \ne 1$, it suffices to prove that
$$(1 + s) \ln\left(1+\frac{2s}{1+s^2}\right)\le 2s$$
or
$$\mathrm{exp}\left(\frac{2s}{1 + s}\right) \ge 1+\frac{2s}{1+s^2}$$
or (using $\mathrm{e}^x \ge 1 + x + \frac12x^2$ for all $x \ge 0$)
$$1 + \frac{2s}{1 + s} + \frac12\left(\frac{2s}{1 + s}\right)^2
\ge 1+\frac{2s}{1+s^2}$$
or
$$\frac{4s^4}{(1 + s)^2(1 + s^2)}\ge 0$$
which is true.
$\phantom{2}$
- If $- 1 < s < 0$, it suffices to prove that
$$-(1 + s) \ln\left(1+\frac{2s}{1+s^2}\right)\le -2s$$
or
$$1+\frac{2s}{1+s^2} \ge \frac{1}{\mathrm{exp}(\frac{-2s}{1+s})}. \tag{3}$$
Using $\mathrm{e}^x \ge 1 + x + \frac12 x^2$ for all $x \ge 0$, we have
$$\frac{1}{\mathrm{exp}(\frac{-2s}{1+s})}
\le \frac{1}{1 + \frac{-2s}{1+s} + \frac12(\frac{-2s}{1+s})^2}
= 1+\frac{2s}{1+s^2}.$$
Thus, (3) is true.
$\phantom{2}$
- If $s < -1$, it suffices to prove that
$$(1 + s) \ln\left(1+\frac{2s}{1+s^2}\right)\le -2s$$
or
$$1+\frac{2s}{1+s^2}\ge \frac{1}{\mathrm{e}(\frac{2s}{1 + s})}$$
or (using $\mathrm{e}^x \ge 1 + x + \frac12 x^2$ for all $x \ge 0$)
$$1+\frac{2s}{1+s^2} \ge \frac{1}{1 + \frac{2s}{1 + s} + \frac12(\frac{2s}{1 + s})^2}$$
or
$$\frac{4s(1 + s)^3}{(1+s^2)(5s^2 + 4s + 1)}\ge 0$$
which is true.
We are done.
